Answer

Homogeneous mixtures of two or more than two components are known as solutions.

There are three types of solutions.

(i) Gaseous solution:

The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.

(ii) Liquid solution:

The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid.

For example, a solution of ethanol in water is a liquid solution.

(iii) Solid solution:

The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

Answer

In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.

Answer

(i) Mole fraction:

The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.

i.e.,

$$ \text{Mole fraction of a component }=\frac{\text { Number of moles of the component }}{\text { Total number of moles of all components }} $$

Mole fraction is denoted by ’ $x$ ‘.

If in a binary solution, the number of moles of the solute and the solvent are $n_{A}$ and $n_{B}$ respectively, then the mole fraction of the solute in the solution is given by,

$x_{A}=\frac{n_{A}}{n_{A}+n_{B}}$

Similarly, the mole fraction of the solvent in the solution is given as:

$x_{B}=\frac{n_{B}}{n_{A}+n_{B}}$

(ii) Molality

Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:

$\text{Molality (m)}=\frac{\text { Moles of solute }}{\text { Mass of solvent in } \mathrm{kg}}$

(iii) Molarity

Molarity $(M)$ is defined as the number of moles of the solute dissolved in one Litre of the solution.

It is expressed as:

Molarity (M) $=\frac{\text { Moles of solute }}{\text { Volume of solution in Litre }}$

(iv) Mass percentage:

The mass percentage of a component of a solution is defined as the mass of the solute in grams present in $100 \mathrm{~g}$ of the solution. It is expressed as:

$$ \text{Mass $\%$ of a component }=\frac{\text { Mass of component in solution }}{\text { Total mass of solution }} \times 100 $$

Answer

Concentrated nitric acid used in laboratory work is $68 \%$ nitric acid by mass in an aqueous solution. This means that $68 \mathrm{~g}$ of nitric acid is dissolved in $100 \mathrm{~g}$ of the solution.

Molar mass of nitric acid $\left(\mathrm{HNO_3}\right)=1 \times 1+1 \times 14+3 \times 16=63 \mathrm{~g} \mathrm{~mol}^{-1}$

Then, number of moles of $\mathrm{HNO_3}=\frac{68}{63} \mathrm{~mol}$

$=1.079 \mathrm{~mol}$

Given,

Density of solution $=1.504 \mathrm{~g} \mathrm{~mL}^{-1}$

$\therefore$ Volume of $100 \mathrm{~g}$ solution $=\frac{100}{1.504} \mathrm{~mL}$

$=66.49 \mathrm{~mL}$

$=66.49 \times 10^{-3} \mathrm{~L}$

Molarity of solution $=\frac{1.079 \mathrm{~mol}}{66.49 \times 10^{-3} \mathrm{~L}}$

$=16.23 \mathrm{M}$

Answer

$10 \% \mathrm{w} / \mathrm{w}$ solution of glucose in water means that $10 \mathrm{~g}$ of glucose in present in $100 \mathrm{~g}$ of the solution i.e., $10 \mathrm{~g}$ of glucose is present in $(100-10) \mathrm{g}=90 \mathrm{~g}$ of water.

Molar mass of glucose $\left(\mathrm{C_6} \mathrm{H_12} \mathrm{O_6}\right)=6 \times 12+12 \times 1+6 \times 16=180 \mathrm{~g} \mathrm{~mol}^{-1}$

$$ \text{Then, number of moles of glucose }=\frac{10}{180} \mathrm{~mol} $$

$=0.056 \mathrm{~mol}$

$\therefore$ Molality of solution $\quad=\frac{0.056 \mathrm{~mol}}{0.09 \mathrm{~kg}}=0.62 \mathrm{~m}$

Number of moles of water $=\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}$

$=5 \mathrm{~mol}$

$$ \begin{aligned} \Rightarrow \text { Mole fraction of glucose } \left(x_{\mathrm{g}}\right) & =\frac{0.056}{0.056+5} \\ \quad & =0.011 \end{aligned} $$

And, mole fraction of water $x_{\mathrm{w}}=1-x_{\mathrm{g}}$

$=1-0.011$

$=0.989$

If the density of the solution is $1.2 \mathrm{~g} \mathrm{~mL}^{-1}$, then the volume of the $100 \mathrm{~g}$ solution can be given as:

$=\frac{100 \mathrm{~g}}{1.2 \mathrm{~g} \mathrm{~mL}^{-1}}$

$=83.33 \mathrm{~mL}$

$=83.33 \times 10^{-3} \mathrm{~L}$

$\therefore$ Molarity of the solution $=\frac{0.056 \mathrm{~mol}}{83.33 \times 10^{-3} \mathrm{~L}}$

$=0.67 \mathrm{M}$

Answer

Let the amount of $\mathrm{Na_2} \mathrm{CO_3}$ in the mixture be $x \mathrm{~g}$.

Then, the amount of $\mathrm{NaHCO_3}$ in the mixture is $(1-x) \mathrm{g}$.

Molar mass of $\mathrm{Na_2} \mathrm{CO_3}=2 \times 23+1 \times 12+3 \times 16$

$=106 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles $\mathrm{Na_2} \mathrm{CO_3}=\frac{x}{106} \mathrm{~mol}$

Molar mass of $\mathrm{NaHCO_3}=1 \times 23+1 \times 1 \times 12+3 \times 16$

$=84 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles of $\mathrm{NaHCO_3}=\frac{1-x}{84} \mathrm{~mol}$

According to the question,

$\frac{x}{106}=\frac{1-x}{84}$

$\Rightarrow 84 x=106-106 x$

$\Rightarrow 190 x=106$

$\Rightarrow x=0.5579$

Therefore, number of moles of $\mathrm{Na_2} \mathrm{CO_3}=\frac{0.5579}{106} \mathrm{~mol}$

$=0.0053 \mathrm{~mol}$

And, number of moles of $\mathrm{NaHCO_3}=\frac{1-0.5579}{84}$ $=0.0053 \mathrm{~mol}$

$\mathrm{HCl}$ reacts with $\mathrm{Na_2} \mathrm{CO_3}$ and $\mathrm{NaHCO_3}$ according to the following equation.

$\underset{\mathrm{2~mol}}{2 \mathrm{HCl}}+ \underset{\mathrm{1~mol}}{\mathrm{Na_2} \mathrm{CO_3}} \longrightarrow 2 \mathrm{NaCl}+\mathrm{H_2} \mathrm{O}+\mathrm{CO_2}$

$\underset{\mathrm{1~mol}}{\mathrm{HCl}}+\underset{\mathrm{1~mol}}{\mathrm{NaHCO_3}} \longrightarrow \mathrm{NaCl}+\mathrm{H_2} \mathrm{O}+\mathrm{CO_2}$

$1 \quad 1 $

1 mol of $\mathrm{Na_2} \mathrm{CO_3}$ reacts with $2 \mathrm{~mol}$ of $\mathrm{HCl}$.

Therefore, $0.0053 \mathrm{~mol}$ of $\mathrm{Na_2} \mathrm{CO_3}$ reacts with $2 \times 0.0053 \mathrm{~mol}=0.0106 \mathrm{~mol}$.

Similarly, 1 mol of $\mathrm{NaHCO_3}$ reacts with $1 \mathrm{~mol}$ of $\mathrm{HCl}$.

Therefore, $0.0053 \mathrm{~mol}$ of $\mathrm{NaHCO_3}$ reacts with $0.0053 \mathrm{~mol}$ of $\mathrm{HCl}$.

Total moles of $\mathrm{HCl}$ required $=(0.0106+0.0053) \mathrm{mol}$

$=0.0159 \mathrm{~mol}$

In $0.1 \mathrm{M}$ of $\mathrm{HCl}$,

$0.1 \mathrm{~mol}$ of $\mathrm{HCl}$ is preset in $1000 \mathrm{~mL}$ of the solution.

Therefore, $0.0159 \mathrm{~mol}$ of $\mathrm{HCl}$ is present in

$$ \frac{1000 \times 0.0159}{0.1} \mathrm{~mol} $$

$=159 \mathrm{~mL}$ of the solution

Hence, $159 \mathrm{~mL}$ of $0.1 \mathrm{M}$ of $\mathrm{HCl}$ is required to react completely with $1 \mathrm{~g}$ mixture of $\mathrm{Na_2} \mathrm{CO_3}$ and $\mathrm{NaHCO_3,}$, containing equimolar amounts of both.

Answer

Total amount of solute present in the mixture is given by,

$300 \times \frac{25}{100}+400 \times \frac{40}{100}$

$=75+160$

$=235 \mathrm{~g}$

Total amount of solution $=300+400=700 \mathrm{~g}$

Therefore, mass percentage (w/w) of the solute in the resulting solution,

$$ =\frac{235}{700} \times 100 \% $$

$=33.57 \%$

And, mass percentage ( $w / w$ ) of the solvent in the resulting solution,

$=(100-33.57) \%$

$=66.43 \%$

Answer

Molar mass of ethylene glycol $\left[\mathrm{C_2} \mathrm{H_4}(\mathrm{OH})_{2}\right]=2 \times 12+6 \times 1+2 \times 16$

$=62 \mathrm{gmol}^{-1}$

Number of moles of ethylene glycol

$$ =\frac{222.6 \mathrm{~g}}{62 \mathrm{gmol}^{-1}} $$

$=3.59 \mathrm{~mol}$

Therefore, molality of the solution

$$ =\frac{3.59 \mathrm{~mol}}{0.200 \mathrm{~kg}} $$

$=17.95 \mathrm{~m}$

Total mass of the solution $=(222.6+200) \mathrm{g}$

$=422.6 \mathrm{~g}$

Given,

Density of the solution $=1.072 \mathrm{~g} \mathrm{~mL}^{-1}$

$\therefore$ Volume of the solution $=\frac{422.6 \mathrm{~g}}{1.072 \mathrm{~g} \mathrm{~mL}^{-1}}$

$=394.22 \mathrm{~mL}$

$=0.3942 \times 10^{-3} \mathrm{~L}$

$\Rightarrow$ Molarity of the solution $=\frac{3.59 \mathrm{~mol}}{0.39422 \times 10^{-3} \mathrm{~L}}$

$=9.11 \mathrm{M}$

Answer

(i) 15 ppm (by mass) means 15 parts per million $\left(10^{6}\right)$ of the solution.

Therefore, percent by mass $=\frac{15}{10^{6}} \times 100 \%$

$=1.5 \times 10^{-3} \%$

(ii) Molar mass of chloroform $\left(\mathrm{CHCl_3}\right)=1 \times 12+1 \times 1+3 \times 35.5$

$=119.5 \mathrm{~g} \mathrm{~mol}^{-1}$

Now, according to the question,

$15 \mathrm{~g}$ of chloroform is present in $10^{6} \mathrm{~g}$ of the solution.

i.e., $15 \mathrm{~g}$ of chloroform is present in $\left(10^{6}-15\right) /10^{6} \mathrm{~g}$ of water.

$\therefore$ Molality of the solution $=\frac{\frac{15}{119.5} \mathrm{~mol}}{10^{6} \times 10^{-3} \mathrm{~kg}}$

$=1.26 \times 10^{-4} \mathrm{~m}$

Answer

In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol-alcohol and water-water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.

Answer

Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.

Gas + Liquid $\longrightarrow$ Solution + Heat

Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.

Answer

Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If $p$ is the partial pressure of the gas in the vapour phase and $x$ is the mole fraction of the gas, then Henry’s law can be expressed as:

$p=K_{\mathrm{H}} x$

Where,

$K_{\mathrm{H}}$ is Henry’s law constant

Some important applications of Henry’s law are mentioned below.

(i) Bottles are sealed under high pressure to increase the solubility of $\mathrm{CO_2}$ in soft drinks and soda water.

(ii) Henry’s law states that the solubility of gases increases with an increase in pressure. Therefore, when a scuba diver dives deep into the sea, the increased sea pressure causes the nitrogen present in air to dissolve in his blood in great amounts. As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading to the formation of nitrogen bubbles in the blood. This results in the blockage of capillaries and leads to a medical condition known as ‘bends’ or ‘decompression sickness’.

Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.

(iii) The concentration of oxygen is low in the blood and tissues of people living at high altitudes such as climbers. This is because at high altitudes, partial pressure of oxygen is less than that at ground level. Low-blood oxygen causes climbers to become weak and disables them from thinking clearly. These are symptoms of anoxia.

Answer

Molar mass of ethane $\left(\mathrm{C_2} \mathrm{H_6}\right)=2 \times 12+6 \times 1$

$=30 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles present in $6.56 \times 10^{-3} \mathrm{~g}$ of ethane $=\frac{6.56 \times 10^{-3}}{30}$

$=2.187 \times 10^{-4} \mathrm{~mol}$

Let the number of moles of the solvent be $x$.

According to Henry’s law,

$$ \begin{aligned} & p=K_{H} X \\ & \Rightarrow 1 \text { bar }=K_{H} \cdot \frac{2.187 \times 10^{-4}}{2.187 \times 10^{-4}+x} \\ & \Rightarrow 1 \text { bar }=K_{H} \cdot \frac{2.187 \times 10^{-4}}{x}\left(\text { Since } x > > 2.187 \times 10^{-4}\right) \\ & \Rightarrow K_{H}=\frac{x}{2.187 \times 10^{-4}} \text { bar } \end{aligned} $$

Answer

According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

Vapour pressure of a two-component solution showing positive deviation from Raoult’s law

Vapour pressure of a two-component solution showing negative deviation from Raoult’s law

In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.

$\Delta_{\text {sol }} H=0$

In the case of solutions showing positive deviations, absorption of heat takes place.

$\therefore \Delta_{\text {sol }} H=$ Positive

In the case of solutions showing negative deviations, evolution of heat takes place.

$\therefore \Delta_{\text {sol }} H=$ Negative

Answer

Here,

Vapour pressure of the solution at normal boiling point $\left(p_{1}\right)=1.004$ bar

Vapour pressure of pure water at normal boiling point $\left(p_{1}^{0}\right)=1.013$ bar

Mass of solute, $\left(w_{2}\right)=2 \mathrm{~g}$

Mass of solvent (water), $\left(w_{1}\right)=98 \mathrm{~g}$

Molar mass of solvent (water), $\left(M_{1}\right)=18 \mathrm{~g} \mathrm{~mol}^{-1}$

According to Raoult’s law,

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$

$\Rightarrow \frac{1.013-1.004}{1.013}=\frac{2 \times 18}{M_{2} \times 98}$

$\Rightarrow \frac{0.009}{1.013}=\frac{2 \times 18}{M_{2} \times 98}$

$\Rightarrow M_{2}=\frac{1.013 \times 2 \times 18}{0.009 \times 98}$

$=41.35 \mathrm{~g} \mathrm{~mol}^{-1}$

Hence, the molar mass of the solute is $41.35 \mathrm{~g} \mathrm{~mol}^{-1}$.

Answer

Vapour pressure of heptane $\left(p_{1}^{0}\right)=105.2 \mathrm{kPa}$

Vapour pressure of octane $\left(p_{2}^{0}\right)=46.8 \mathrm{kPa}$

We know that,

Molar mass of heptane $\left(\mathrm{C_7} \mathrm{H_16}\right)=7 \times 12+16 \times 1$

$=100 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles of heptane $=\frac{26}{100} \mathrm{~mol}$

$=0.26 \mathrm{~mol}$

Molar mass of octane $\left(\mathrm{C_8} \mathrm{H_18}\right)=8 \times 12+18 \times 1$

$=114 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles of octane $=\frac{35}{114} \mathrm{~mol}$

$=0.31 \mathrm{~mol}$

Mole fraction of heptane, $x_{1}=\frac{0.26}{0.26+0.31}$

$=0.456$

And, mole fraction of octane, $x_{2}=1-0.456$

$=0.544$

Now, partial pressure of heptane, $p_{1}=x_{1} p_{1}^{0}$

$=0.456 \times 105.2$

$=47.97 \mathrm{kPa}$

Partial pressure of octane, $p_{2}=x_{2} p_{2}^{0}$

$=0.544 \times 46.8$

$=25.46 \mathrm{kPa}$

Hence, vapour pressure of solution, $p_{\text {total }}=p_{1}+p_{2}$

$=47.97+25.46$

$=73.43 \mathrm{kPa}$

Answer

1 molal solution means $1 \mathrm{~mol}$ of the solute is present in $1000 \mathrm{~g}$ of the solvent (water).

Molar mass of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ Number of moles present in $1000 \mathrm{~g}$ of water $=\frac{1000}{18}$

$=55.56 \mathrm{~mol}$

Therefore, mole fraction of the solute in the solution is

$x_{2}=\frac{1}{1+55.56}=0.0177$

It is given that,

Vapour pressure of water, $p_{1}^{0}=12.3 \mathrm{kPa}$

Applying the relation, $\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=x_{2}$

$\Rightarrow \frac{12.3-p_{1}}{12.3}=0.0177$

$\Rightarrow 12.3-p_{1}=0.2177$

$\Rightarrow p_{1}=12.0823$

$=12.08 \mathrm{kPa}$ (approximately)

Hence, the vapour pressure of the solution is $12.08 \mathrm{kPa}$.

Answer

Let the vapour pressure of pure octane be $p_{1}^{0}$.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is $\frac{80}{100} p_{1}^{0}=0.8 p_{1}^{0}$.

Molar mass of solute, $M_{2}=40 \mathrm{~g} \mathrm{~mol}^{-1}$

Mass of octane, $w_{1}=114 \mathrm{~g}$

Molar mass of octane, $\left(\mathrm{C_8} \mathrm{H_18}\right), M_{1}=8 \times 12+18 \times 1$

$=114 \mathrm{~g} \mathrm{~mol}^{-1}$

Applying the relation,

$ \begin{aligned} & \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{w_{2} \times M_{1}}{M_{2} \times w_{1}} \\ & \Rightarrow \frac{p_{1}^{0}-0.8 p_{1}^{0}}{p_{1}^{0}}=\frac{w_{2} \times 114}{40 \times 114} \end{aligned} $

$\Rightarrow \frac{0.2 p_{1}^{0}}{p_{1}^{0}}=\frac{w_{2}}{40}$

$\Rightarrow 0.2=\frac{w_{2}}{40}$

$\Rightarrow w_{2}=8 \mathrm{~g}$

Hence, the required mass of the solute is $8 \mathrm{~g}$.

Answer

(i) Let, the molar mass of the solute be $\mathrm{M} \mathrm{g} \mathrm{mol}^{-1}$

$$ n_{1}=\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=5 \mathrm{~mol} $$

Now, the no. of moles of solvent (water),

$$ n_{2}=\frac{30 \mathrm{~g}}{\mathrm{M} \mathrm{mol}^{-1}}=\frac{30}{\mathrm{M}} \mathrm{mol} $$

And, the no. of moles of solute,

$$ p_{1}=2.8 \mathrm{kPa} $$

Applying the relation:

$$ \begin{align*} & \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}} \\ \\ & \Rightarrow \frac{p_{1}^{0}-2.8}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{5+\frac{30}{\mathrm{M}}} \\ \\ & \Rightarrow 1-\frac{2.8}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}+30}{\mathrm{M}} \\ \\ & \Rightarrow 1-\frac{2.8}{p_{1}^{0}}=\frac{30}{5 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{2.8}{p_{1}^{0}}=1-\frac{30}{5 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{2.8}{p_{1}^{0}}=\frac{5 \mathrm{M}+30-30}{5 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{2.8}{p_{1}^{0}}=\frac{5 \mathrm{M}}{5 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{p^{0}}{2.8} =\frac{5 \mathrm{M}+30}{5 \mathrm{M}} \tag{i} \\ \\ \end{align*} $$

After the addition of $18 \mathrm{~g}$ of water:

$$ \begin{aligned} & n_{1}=\frac{90+18 \mathrm{~g}}{18}=6 \mathrm{~mol} \\ & p_{1}=2.9 \mathrm{kPa} \end{aligned} $$

Again, applying the relation:

$$ \begin{align*} & \frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{2}}{n_{1}+n_{2}} \\ \\ & \Rightarrow \frac{p_{1}^{0}-2.9}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{6+\frac{30}{\mathrm{M}}} \\ \\ & \Rightarrow 1-\frac{2.9}{p_{1}^{0}}=\frac{\frac{30}{\mathrm{M}}}{\frac{6 \mathrm{M}+30}{\mathrm{M}}} \\ \\ & \Rightarrow 1-\frac{2.9}{p_{1}^{0}}=\frac{30}{6 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{2.9}{p_{1}^{0}}=1-\frac{30}{6 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{2.9}{p_{1}^{0}}=\frac{6 \mathrm{M}+30-30}{6 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{2.9}{p_{1}^{0}}=\frac{6 \mathrm{M}}{6 \mathrm{M}+30} \\ \\ & \Rightarrow \frac{p_{1}^{0}}{2.9}=\frac{6 \mathrm{M}+30}{6 \mathrm{M}} \tag{ii} \end{align*} $$

Dividing equation (i)by (ii), we have:

$$ \begin{aligned} & \frac{2.9}{2.8}=\frac{\frac{5 \mathrm{M}+30}{5 \mathrm{M}}}{\frac{6 \mathrm{M}+30}{6 \mathrm{M}}} \\ \\ & \Rightarrow \frac{2.9}{2.8} \times \frac{6 \mathrm{M}+30}{6}=\frac{5 \mathrm{M}+30}{5} \\ \\ & \Rightarrow 2.9 \times 5 \times(6 \mathrm{M}+30)=2.8 \times 6 \times(5 \mathrm{M}+30) \\ \\ & \Rightarrow 87 \mathrm{M}+435=84 \mathrm{M}+504 \\ \\ & \Rightarrow 3 \mathrm{M}=69 \\ \\ & \Rightarrow \mathrm{M}=23 \mathrm{u} \end{aligned} $$

Therefore, the molar mass of the solute is $23 \mathrm{~g} \mathrm{~mol}^{-1}$.

(ii) Putting the value of ‘M’ in equation (i), we have:

$$ \begin{aligned} & \frac{p_{1}^{0}}{2.8}=\frac{5 \times 23+30}{5 \times 23} \\ \\ & \Rightarrow \frac{p_{1}^{0}}{2.8}=\frac{145}{115} \\ \\ & \Rightarrow p_{1}^{0}=3.53 \end{aligned} $$

Hence, the vapour pressure of water at $298 \mathrm{~K}$ is $3.53 \mathrm{kPa}$.

Answer

Here, $\Delta T_{f}=(273.15-271) \mathrm{K}$

$=2.15 \mathrm{~K}$

Molar mass of sugar $\left(\mathrm{C_12} \mathrm{H_22} \mathrm{O_11}\right)=12 \times 12+22 \times 1+11 \times 16$

$=342 \mathrm{~g} \mathrm{~mol}^{-1}$

$5 \%$ solution (by mass) of cane sugar in water means $5 \mathrm{~g}$ of cane sugar is present in $(100-5) \mathrm{g}=95 \mathrm{~g}$ of water.

Now, number of moles of cane sugar

$$ =\frac{5}{342} \mathrm{~mol} $$

$=0.0146 \mathrm{~mol}$

Therefore, molality of the solution,

$$ m=\frac{0.0146 \mathrm{~mol}}{0.095 \mathrm{~kg}} $$

$=0.1537 \mathrm{~mol} \mathrm{~kg}^{-1}$

Applying the relation,

$\Delta T_{f}=K_{f} \times m$

$ \begin{aligned} \Rightarrow K_{f} & =\frac{\Delta T_{f}}{m} \\ & =\frac{2.15 \mathrm{~K}}{0.1537 \mathrm{~mol} \mathrm{~kg}^{-1}} \end{aligned} $

$=13.99 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

Molar of glucose $\left(\mathrm{C_6} \mathrm{H_12} \mathrm{O_6}\right)=6 \times 12+12 \times 1+6 \times 16$

$=180 \mathrm{~g} \mathrm{~mol}^{-1}$

$5 \%$ glucose in water means $5 \mathrm{~g}$ of glucose is present in $(100-5) \mathrm{g}=95 \mathrm{~g}$ of water.

$\therefore$ Number of moles of glucose $=\frac{5}{180} \mathrm{~mol}$

$=0.0278 \mathrm{~mol}$

Therefore, molality of the solution,

$$ m=\frac{0.0278 \mathrm{~mol}}{0.095 \mathrm{~kg}} $$

$=0.2926 \mathrm{~mol} \mathrm{~kg}^{-1}$

Applying the relation,

$\Delta T_{f}=K_{f} \times m$

$=13.99 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.2926 \mathrm{~mol} \mathrm{~kg}^{-1}$

$=4.09 \mathrm{~K}$ (approximately)

Hence, the freezing point of $5 \%$ glucose solution is ($273.15$ - 4.09) K= $269.06 \mathrm{~K}$.

Answer

We know that, $M_{2}=\frac{1000 \times w_{2} \times k_{f}}{\Delta T_{f} \times w_{1}}$

Then, $M_{\mathrm{AB_2}}=\frac{1000 \times 1 \times 5.1}{2.3 \times 20}$

$=110.87 \mathrm{~g} \mathrm{~mol}^{-1}$

$M_{\mathrm{AB_4}}=\frac{1000 \times 1 \times 5.1}{1.3 \times 20}$

$=196.15 \mathrm{~g} \mathrm{~mol}^{-1}$

Now, we have the molar masses of $\mathrm{AB_2}$ and $\mathrm{AB_4}$ as $110.87 \mathrm{~g} \mathrm{~mol}^{-1}$ and $196.15 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.

Let the atomic masses of $A$ and $B$ be $x$ and $y$ respectively.

Now, we can write:

$x+2 y=110.87\quad \quad \quad \quad \text{(i)}$

$x+4 y=196.15\quad \quad \quad \quad \text{(ii)}$

Subtracting equation (i) from (ii), we have

$2 y=85.28$

$\Rightarrow y=42.64$

Putting the value of ’ $y$ ’ in equation (1), we have

$x+2 \times 42.64=110.87$

$\Rightarrow x=25.59$

Hence, the atomic masses of $A$ and $B$ are $25.59 \mathrm{u}$ and $42.64 \mathrm{u}$ respectively.

Answer

Here,

$T=300 \mathrm{~K}$

$\pi=1.52$ bar

$\mathrm{R}=0.083$ bar L K ${ }^{-1} \mathrm{~mol}^{-1}$

Applying the relation,

$\pi=C R T$

$ \begin{aligned} \Rightarrow C & =\frac{\pi}{\mathrm{R} T} \\ & =\frac{1.52 \mathrm{bar}}{0.083 \mathrm{bar} \mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}} \end{aligned} $

$=0.061 \mathrm{~mol}$

Since the volume of the solution is $1 \mathrm{~L}$, the concentration of the solution would be $0.061 \mathrm{M}$.

Answer

(i) Van der Wall’s forces of attraction.

(ii) Van der Wall’s forces of attraction.

(iii) Ion-diople interaction.

(iv) Dipole-dipole interaction.

(v) Dipole-dipole interaction.

Answer

$n$-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the $n$-octane.

The order of increasing polarity is:

Cyclohexane $<\mathrm{CH_3} \mathrm{CN}<\mathrm{CH_3} \mathrm{OH}<\mathrm{KCl}$

Therefore, the order of increasing solubility is:

$\mathrm{KCl}<\mathrm{CH_3} \mathrm{OH}<\mathrm{CH_3} \mathrm{CN}<$ Cyclohexane

Answer

(i) Phenol $\left(\mathrm{C_6} \mathrm{H_5} \mathrm{OH}\right)$ has the polar group $-\mathrm{OH}$ and non-polar group $-\mathrm{C_6} \mathrm{H_5}$. Thus, phenol is partially soluble in water.

(ii) Toluene $\left(\mathrm{C_6} \mathrm{H_5}-\mathrm{CH_3}\right)$ has no polar groups. Thus, toluene is insoluble in water.

(iii) Formic acid $(\mathrm{HCOOH})$ has the polar group - $\mathrm{OH}$ and can form $\mathrm{H}$-bond with water. Thus, formic acid is highly soluble in water.

(iv) Ethylene glycol has polar - $\mathrm{OH}$ group and can form $\mathrm{H}$-bond. Thus, it is highly soluble in water.

(v) Chloroform is insoluble in water.

(vi) Pentanol $\left(\mathrm{C_5} \mathrm{H_11} \mathrm{OH}\right)$ has polar $-\mathrm{OH}$ group, but it also contains a very bulky non-polar $-\mathrm{C_5} \mathrm{H_11}$ group. Thus, pentanol is partially soluble in water.

Answer

Number of moles present in $92 \mathrm{~g}$ of $\mathrm{Na}^{+}$ions $=\frac{92 \mathrm{~g}}{23 \mathrm{~g} \mathrm{~mol}^{-1}}$

$=4 \mathrm{~mol}$

Therefore, molality of $\mathrm{Na}^{+}$ions in the lake $=\frac{4 \mathrm{~mol}}{1 \mathrm{~kg}}$

$=4 \mathrm{~m}$

Answer

Solubility product of CuS, $K_{\mathrm{sp}}=6 \times 10^{-16}$

Let $s$ be the solubility of CuS in $\mathrm{mol}^{-1}$.

$\mathrm{CuS} \leftrightarrow \mathrm{Cu}^{2+}+\mathrm{S}^{2-}$

Now, $K_{s p}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{S}^{2-}\right]$

$=s \times s$

$=s^{2}$

Then, we have, $K_{\mathrm{sp}}=s^{2}=6 \times 10^{-16}$

$\Rightarrow s=\sqrt{6 \times 10^{-16}}$

$=2.45 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}$

Hence, the maximum molarity of CuS in an aqueous solution is $2.45 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}$.

Answer

$6.5 \mathrm{~g}$ of $\mathrm{C_9} \mathrm{H_8} \mathrm{O_4}$ is dissolved in $450 \mathrm{~g}$ of $\mathrm{CH_3} \mathrm{CN}$.

Then, total mass of the solution $=(6.5+450) \mathrm{g}$

$=456.5 \mathrm{~g}$

Therefore, mass percentage of $\mathrm{C_9} \mathrm{H_8} \mathrm{O_4}$

$$ =\frac{6.5}{456.5} \times 100 \% $$

$=1.424 \%$

Answer

The molar mass of nalorphene $\left(\mathrm{C_19} \mathrm{H_21} \mathrm{NO_3}\right)$ is given as:

$19 \times 12+21 \times 1+1 \times 14+3 \times 16=311 \mathrm{~g} \mathrm{~mol}^{-1}$

In $1.5 \times 10^{-3} \mathrm{~m}$ aqueous solution of nalorphene,

$1 \mathrm{~kg}(1000 \mathrm{~g})$ of water contains $1.5 \times 10^{-3} \mathrm{~mol}=1.5 \times 10^{-3} \times 311 \mathrm{~g}$

$=0.4665 \mathrm{~g}$

Therefore, total mass of the solution $=(1000+0.4665) \mathrm{g}$

$=1000.4665 \mathrm{~g}$

This implies that the mass of the solution containing $0.4665 \mathrm{~g}$ of nalorphene is $1000.4665 \mathrm{~g}$.

Therefore, mass of the solution containing $1.5 \mathrm{mg}$ of nalorphene is:

$$ \begin{aligned} & \frac{1000.4665 \times 1.5 \times 10^{-3}}{0.4665} \mathrm{~g} \\ & =3.22 \mathrm{~g} \end{aligned} $$

Hence, the mass of aqueous solution required is $3.22 \mathrm{~g}$.

Note:There is a slight variation in this answer and the one given in the NCERT textbook.

Answer

$0.15 \mathrm{M}$ solution of benzoic acid in methanol means,

$1000 \mathrm{~mL}$ of solution contains $0.15 \mathrm{~mol}$ of benzoic acid

Therefore, $250 \mathrm{~mL}$ of solution contains $=\frac{0.15 \times 250}{1000}$ mol of benzoic acid

$=0.0375 \mathrm{~mol}$ of benzoic acid

Molar mass of benzoic acid $\left(\mathrm{C_6} \mathrm{H_5} \mathrm{COOH}\right)=7 \times 12+6 \times 1+2 \times 16$

$=122 \mathrm{~g} \mathrm{~mol}^{-1}$

Hence, required benzoic acid $=0.0375 \mathrm{~mol} \times 122 \mathrm{~g} \mathrm{~mol}^{-1}$

$=4.575 \mathrm{~g}$

Answer

Among $\mathrm{H}, \mathrm{Cl}$, and $\mathrm{F}, \mathrm{H}$ is least electronegative while $\mathrm{F}$ is most electronegative. Then, $\mathrm{F}$ can withdraw electrons towards itself more than $\mathrm{Cl}$ and $\mathrm{H}$. Thus, trifluoroacetic acid can easily lose $\mathrm{H}^{+}$ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:

Acetic acid < trichloroacetic acid < trifluoroacetic acid

Answer

Molar mass of $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH}=15+14+13+35.5+12+16+16+1$

$=122.5 \mathrm{~g} \mathrm{~mol}^{-1}$

$\therefore$ No. of moles present in $10 \mathrm{~g}$ of

$$ \mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH}=\frac{10 \mathrm{~g}}{122.5 \mathrm{~g} \mathrm{~mol}^{-1}} $$

$=0.0816 \mathrm{~mol}$

It is given that $10 \mathrm{~g}$ of

$\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH}$ is added to $250 \mathrm{~g}$ of water.

$\therefore$ Molality of the solution,

$$ =\frac{0.0186}{250} \times 1000 $$

$=0.3264 \mathrm{~mol} \mathrm{~kg}^{-1}$

Let $a$ be the degree of dissociation of $\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH}$.

$\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH}$ undergoes dissociation according to the following equation:

$ \begin{array}{lcccc} & \mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH} &\leftrightarrow &\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOO}^{-}& +\mathrm{H}^{+} \\ \text{Initial conc. } & \mathrm{C} \text{mol L}^{-1} & & 0 & 0 \\ \text{At equilibrium } & \mathrm{C1-\alpha} & & \mathrm{C\alpha} & \mathrm{C\alpha} \end{array} $

$ \begin{aligned} & \therefore K_{a}=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)} \\ & =\frac{C \alpha^{2}}{1-\alpha} \end{aligned} $

Since $\alpha$ is very small with respect to $1,1-\alpha$

Now,

$ K_{a}=\frac{C \alpha^{2}}{1} $

$\Rightarrow K_{a}=C \alpha^{2}$

$\Rightarrow \alpha=\sqrt{\frac{K_{a}}{C}}$

$$ \begin{aligned} & =\sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} \quad\left(\because K_{a}=1.4 \times 10^{-3}\right) \\ & =0.0655 \end{aligned} $$

Again,

$ \begin{array}{lcccc} & \mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOOH} &\leftrightarrow &\mathrm{CH_3} \mathrm{CH_2} \mathrm{CHClCOO}^{-}& +\mathrm{H}^{+} \\ \text{Initial conc. } & 1 & & 0 & 0 \\ \text{At equilibrium } & \mathrm{1-\alpha} & & \mathrm{\alpha} & \mathrm{\alpha} \end{array} $

Total moles of equilibrium $=1-\alpha+\alpha+\alpha$

$$ \begin{aligned} & =1+\alpha \\ & \therefore i=\frac{1+\alpha}{1} \\ & =1+\alpha \\ & =1+0.0655 \\ & =1.0655 \end{aligned} $$

Hence, the depression in the freezing point of water is given as:

$$ \begin{aligned} & \Delta T_{f}=i . K_{f} m \\ & =1.0655 \times 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.3264 \mathrm{~mol} \mathrm{~kg}^{-1} \\ & =0.65 \mathrm{~K} \end{aligned} $$

Answer

It is given that:

$w_{1}=500 \mathrm{~g}$

$w_{2}=19.5 \mathrm{~g}$

$K_{f}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

$\Delta T_{f}=1 \mathrm{~K}$

We know that:

$M_{2}=\frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}}$

$=\frac{1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 19.5 \mathrm{~g} \times 1000 \mathrm{~g} \mathrm{~kg}^{-1}}{500 \mathrm{~g} \times 1 \mathrm{~K}}$

$=72.54 \mathrm{~g} \mathrm{~mol}^{-1}$

Therefore, observed molar mass of $\mathrm{CH_2} \mathrm{FCOOH},\left(M_{2}\right)_{\text {obs }}=72.54 \mathrm{~g} \mathrm{~mol}$

The calculated molar mass of $\mathrm{CH_2} \mathrm{FCOOH_\text {is: }}$

$\left(M_{2}\right)_{\text {cal }}=14+19+12+16+16+1$

$=78 \mathrm{~g} \mathrm{~mol}^{-1}$

$=\frac{78 \mathrm{~g} \mathrm{~mol}^{-1}}{72.54 \mathrm{~g} \mathrm{~mol}^{-1}}$

$=1.0753$

Let abe the degree of dissociation of $\mathrm{CH_2} \mathrm{FCOOH}$

$ \begin{array}{lcccc} & \mathrm{CH_2} \mathrm{FCOOH} &\leftrightarrow &\mathrm{CH_2} \mathrm{FCOO}^{-}& +\mathrm{H}^{+} \\ \text{Initial conc. } & \mathrm{C} \text{mol L}^{-1} & & 0 & 0 \\ \text{At equilibrium } & \mathrm{C1-\alpha} & & \mathrm{C\alpha} & \mathrm{C\alpha} \end{array} $

$\text { Total }=\mathrm{C}(1+\alpha)$

$$ \begin{aligned} & \therefore i=\frac{C(1+\alpha)}{C} \\ \\ & \Rightarrow i=1+\alpha \\ \\ & \Rightarrow \alpha=i-1 \\ \\ & =1.0753-1 \\ \\ & =0.0753 \end{aligned} $$

Now, the value of $K_{a}$ is given as:

$$ \begin{aligned} K_{a} & =\frac{\left[\mathrm{CH_2} \mathrm{FCOO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH_2} \mathrm{FCOOH}\right]} \\ \\ = & \frac{C \alpha \cdot C \alpha}{C(1-\alpha)} \\ \\ = & \frac{C \alpha^{2}}{1-\alpha} \end{aligned} $$

Taking the volume of the solution as $500 \mathrm{~mL}$, we have the concentration:

$$ \begin{aligned} & C=\frac{\frac{19.5}{78}}{500} \times 1000 \mathrm{M} \\ \\ & =0.5 \mathrm{M} \\ \\ & \qquad K_{a}=\frac{C \alpha^{2}}{1-\alpha} \\ \\ & \text { Therefore, } \\ \\ & =\frac{0.5 \times(0.0753)^{2}}{1-0.0753} \\ \\ & =\frac{0.5 \times 0.00567}{0.9247} \\ \\ & =0.00307 \text { (approximately) } \\ \\ & =3.07 \times 10^{-3} \end{aligned} $$

Answer

Vapour pressure of water, $p_{1}^{0}=17.535 \mathrm{~mm}$ of $\mathrm{Hg}$

Mass of glucose, $w_{2}=25 \mathrm{~g}$

Mass of water, $w_{1}=450 \mathrm{~g}$

We know that,

Molar mass of glucose $\left(\mathrm{C_6} \mathrm{H_12} \mathrm{O_6}\right), M_{2}=6 \times 12+12 \times 1+6 \times 16$

$=180 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of water, $M_{1}=18 \mathrm{~g} \mathrm{~mol}^{-1}$

Then, number of moles of glucose, $\quad n_{2}=\frac{25}{180 \mathrm{~g} \mathrm{~mol}^{-1}}$

$=0.139 \mathrm{~mol}$

And, number of moles of water,

$$ n_{1}=\frac{450 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}} $$

$=25 \mathrm{~mol}$

We know that,

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{n_{1}}{n_{2}+n_{1}}$

$\Rightarrow \frac{17.535-p_{1}}{17.535}=\frac{0.139}{0.139+25}$

$\Rightarrow 17.535-p_{1}=\frac{0.139 \times 17.535}{25.139}$

$\Rightarrow 17.535-p_{1}=0.097$

$\Rightarrow p_{1}=17.44 \mathrm{~mm}$ of $\mathrm{Hg}$

Hence, the vapour pressure of water is $17.44 \mathrm{~mm}$ of $\mathrm{Hg}$.

Answer

Here,

$p=760 \mathrm{~mm} \mathrm{Hg}$

$\mathrm{k_\mathrm{H}}=4.27 \times 10^{5} \mathrm{~mm} \mathrm{Hg}$

According to Henry’s law,

$p=\mathrm{k_\mathrm{H}} \mathrm{X}$

$\Rightarrow x=\frac{p}{k_{\mathrm{H}}}$

$ =\frac{760 \mathrm{~mm} \mathrm{Hg}}{4.27 \times 10^{5} \mathrm{~mm} \mathrm{Hg}} $

$=177.99 \times 10^{-5}$

$=178 \times 10^{-5}$ (approximately)

Hence, the mole fraction of methane in benzene is $178 \times 10^{-5}$.

Answer

Number of moles of liquid A, $\quad n_{\mathrm{A}}=\frac{100}{140} \mathrm{~mol}$

$=0.714 \mathrm{~mol}$

Number of moles of liquid $B$,

$$ n_{\mathrm{B}}=\frac{1000}{180} \mathrm{~mol} $$

$=5.556 \mathrm{~mol}$

Then, mole fraction of $\mathrm{A}$,

$$ x_{\mathrm{A}}=\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}} $$

$=\frac{0.714}{0.714+5.556}$

$=0.114$

And, mole fraction of $\mathrm{B}, x_{\mathrm{B}}=1-0.114$

$=0.886$

Vapour pressure of pure liquid $\mathrm{B}, p_{\mathrm{B}}^{0}=500$ torr

Therefore, vapour pressure of liquid $B$ in the solution,

$$ p_{\mathrm{B}}=p_{\mathrm{B}}^{0} x_{\mathrm{B}} $$

$=500 \times 0.886$ $=443$ torr

Total vapour pressure of the solution, $p_{\text {total }}=475$ torr

$\therefore$ Vapour pressure of liquid $\mathrm{A}$ in the solution,

$p_{\mathrm{A}}=p_{\text {total }}-p_{\mathrm{B}}$

$=475-443$

$=32$ torr

Now,

$ \begin{aligned} p_{\mathrm{A}} & =p_{\mathrm{A}}^{0} x_{\mathrm{A}} \\ \\ \Rightarrow p_{\mathrm{A}}^{0} & =\frac{p_{\mathrm{A}}}{x_{\mathrm{A}}} \\ \\ & =\frac{32}{0.114} \end{aligned} $

$=280.7$ torr

Hence, the vapour pressure of pure liquid A is 280.7 torr.

Answer

From the question, we have the following data

It can be observed from the graph that the plot for the $p_{\text {total }}$ of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Answer

Molar mass of benzene $\left(\mathrm{C_6} \mathrm{H_6}\right)=6 \times 12+6 \times 1$

$=78 \mathrm{~g} \mathrm{~mol}^{-1}$

Molar mass of toluene $\left(\mathrm{C_6} \mathrm{H_5} \mathrm{CH_3}\right)=7 \times 12+8 \times 1$

$=92 \mathrm{~g} \mathrm{~mol}^{-1}$

Now, no. of moles present in $80 \mathrm{~g}$ of benzene $=\frac{80}{78} \mathrm{~mol}=1.026 \mathrm{~mol}$

And, no. of moles present in $100 \mathrm{~g}$ of toluene $=\frac{100}{92} \mathrm{~mol}=1.087 \mathrm{~mol}$

$\therefore$ Mole fraction of benzene, $x_{b}=\frac{1.026}{1.026+1.087}=0.486$

And, mole fraction of toluene, $x_{t}=1-0.486=0.514$

It is given that vapour pressure of pure benzene, $p_{b}^{0}=50.71 \mathrm{~mm} \mathrm{Hg}$

And, vapour pressure of pure toluene, $p_{t}^{0}=32.06 \mathrm{~mm} \mathrm{Hg}$

Therefore, partial vapour pressure of benzene, $p_{b}=x_{b} \times p_{b}$

$=0.486 \times 50.71$

$=24.645 \mathrm{~mm} \mathrm{Hg}$

And, partial vapour pressure of toluene, $p_{t}=x_{t} \times p_{t}$

$=0.514 \times 32.06$

$=16.479 \mathrm{~mm} \mathrm{Hg}$

Hence, mole fraction of benzene in vapour phase is given by:

$ \begin{aligned} & \frac{p_{b}}{p_{b}+p_{t}} \\ & =\frac{24.645}{24.645+16.479} \\ & =\frac{24.645}{41.124} \\ & =0.599 \\ & =0.6 \end{aligned} $

Answer

Percentage of oxygen $\left(\mathrm{O_2}\right)$ in air $=20 \%$

Percentage of nitrogen $\left(\mathrm{N_2}\right)$ in air $=79 \%$

Also, it is given that water is in equilibrium with air at a total pressure of $10 \mathrm{~atm}$, that is, $(10 \times 760) \mathrm{mm} \mathrm{Hg}=7600 \mathrm{~mm}$ $\mathrm{Hg}$

Therefore,

Partial pressure of oxygen, $p_{\mathrm{O_2}}=\frac{20}{100} \times 7600 \mathrm{~mm} \mathrm{Hg}$

$=1520 \mathrm{~mm} \mathrm{Hg}$

Partial pressure of nitrogen,

$$ p_{\mathrm{N_2}}=\frac{79}{100} \times 7600 \mathrm{mmHg} $$

$=6004 \mathrm{mmHg}$

Now, according to Henry’s law:

$p=K_{\mathrm{H}} \cdot X$

For oxygen:

$$ \begin{aligned} p_{\mathrm{O_2}}= & K_{\mathrm{H}} \cdot x_{\mathrm{O_2}} \\ \Rightarrow x_{\mathrm{O_2}} & =\frac{p_{\mathrm{O_2}}}{K_{\mathrm{H}}} \\ & \left.=\frac{1520 \mathrm{~mm} \mathrm{Hg}}{3.30 \times 10^{7} \mathrm{~mm} \mathrm{Hg}} \quad \text { (Given } K_{\mathrm{H}}=3.30 \times 10^{7} \mathrm{~mm} \mathrm{Hg}\right) \\ & =4.61 \times 10^{-5} \end{aligned} $$

For nitrogen:

$$ \begin{aligned} p_{\mathrm{N_2}}= & K_{\mathrm{H}} \cdot x_{\mathrm{N_2}} \\ \Rightarrow x_{\mathrm{N_2}} & =\frac{p_{\mathrm{N_2}}}{K_{\mathrm{H}}} \\ & =\frac{6004 \mathrm{~mm} \mathrm{Hg}}{6.51 \times 10^{7} \mathrm{~mm} \mathrm{Hg}} \\ & =9.22 \times 10^{-5} \end{aligned} $$

Hence, the mole fractions of oxygen and nitrogen in water are $4.61 \times 10^{-5}$ and $9.22 \times 10^{-5}$ respectively.

Answer

We know that,

$\pi=i \frac{n}{V} \mathrm{R} T$

$\Rightarrow \pi=i \frac{w}{M V} \mathrm{R} T$

$\Rightarrow w=\frac{\pi M V}{i \mathrm{R} T}$ $\pi=0.75 \mathrm{~atm}$

$V=2.5 \mathrm{~L}$

$i=2.47$

$T=(27+273) \mathrm{K}=300 \mathrm{~K}$

Here,

$\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$M=1 \times 40+2 \times 35.5$

$=111 \mathrm{~g} \mathrm{~mol}^{-1}$

Therefore, $w=\frac{0.75 \times 111 \times 2.5}{2.47 \times 0.0821 \times 300}$

$=3.42 \mathrm{~g}$

Hence, the required amount of $\mathrm{CaCl_2}$ is $3.42 \mathrm{~g}$.

Answer

When $\mathrm{K_2} \mathrm{SO_4}$ is dissolved in water, $\mathrm{K}^{+}$and $\mathrm{SO_4}^{2-}$ ions are produced.

$\mathrm{K_2} \mathrm{SO_4} \longrightarrow 2 \mathrm{~K}^{+}+\mathrm{SO_4}^{2-}$

Total number of ions produced $=3$

$\therefore i=3$

Given,

$w=25 \mathrm{mg}=0.025 \mathrm{~g}$

$V=2 \mathrm{~L}$

$T=25^{\circ} \mathrm{C}=(25+273) \mathrm{K}=298 \mathrm{~K}$

Also, we know that:

$\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

$M=(2 \times 39)+(1 \times 32)+(4 \times 16)=174 \mathrm{~g} \mathrm{~mol}^{-1}$

Appling the following relation,

$ \begin{aligned} \pi & =i \frac{n}{v} \mathrm{R} T \\ & =i \frac{w}{M} \frac{1}{v} \mathrm{R} T \\ & =3 \times \frac{0.025}{174} \times \frac{1}{2} \times 0.0821 \times 298 \\ & =5.27 \times 10^{-3} \mathrm{~atm} \end{aligned} $