Large quantity - solvent

Other components - solutes

Composite-

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

• 10% glucose in water
  • 10 g of glucose + 90 g of water = 100g 10% aquous glucose solution

Volume percentage = $\frac {\text {Volume of the component in the solution}}{\text {Total volume of the solution}} \times 100$

• 10% ethanol solution in water
  • 10ml of ethanol + enough water = 100ml of solution

Mass by volume $=\frac{\text{ mass of the component in the solution}}{\text {Total volume of the solution}} \times 100$

• 10% ethanol solution in water
  • 10ml of ethanol + enough water = 100ml of solution

PPM(part per million)

= $\frac {\text {parts of the component}}{\text {Total parts}} \times 10^6$

$6 \times 10^{-3}g$ at $O_2$ in 1L water (1030g)

$\frac {6 \times 10^{-3}}{1030g} \times 10^6 = 5.8 PPM$

= 5.8g of $O_2 10^6g$ water

= 5.8$\mu$ g of $O_2$ 1g water

mole fraction

$X_A=\frac{n_A}{n_A+n_B} \quad X_B = \frac{n_B}{n_A+n_B}$

$X_A+X_B=1$

$X_l=\frac{n_l}{\sum_j n_j}$

$\sum_l X_l=12$

molarity = $\frac {\text {moles of solute}}{\text {volume of solution (litre)}}$

0.25m NaCl solution in water

0.25 moles of NaCl in 1L solution

0.25 moles

molality = $\frac {\text {moles of solute}}{\text {weight of solvent (kg)}}$

1m KBr solution in water

1mole of KBr in 1kg of water

mass parcentage = $\frac {\text{mass of the component}}{\text{Total mass}} \times 100$

$=\frac{22}{122+22} \times 100=15.3$

$100-153=84.7$

$\frac{122}{144} \times 100$

$C_6 H_6(72+6=78) \quad =\frac {22g}{78} = n_1 = 0.28$

$CCl_4(12+35.5 \times 4 = 154) \quad W_2= \frac {122g}{154} = n_2$

$\begin{aligned} & X_1=\frac{n_1}{n_1+n_2}=\frac{0.28}{0.28+0.79}= 0.26 \\ & X_2=1-X_1=74 \ & \end{aligned}$

molarity = $\frac{\text{moles of solute}}{\text{volume of solution (L)}}$

a. $\frac {30}{290.93}g \quad Co(NO_3) . 6H_2O \quad \frac{0.103}{4.3} \quad 0.024M$

$V = 4.3L$

$n_2 = 0.103$

molarity = $\frac {\text{moles of solute}}{\text{volume of solution(Litre)}} \quad =\frac {0.5 \times 30}{1000 \times \frac {500}{1000}}$

a. $30ml 0.5 M H_2SO_4 \rightarrow 500ml$

$0.5 = \frac {\text{moles of solute}}{30/1000}$

moles of solute = $\frac {0.5 \times 30}{1000}$

molality = $\frac {\text{moles of solute}}{\text {Wt of solvent}}$

0.25 =

2.5kg of solution

0.25m

$Urea(NH_2CO NH_2) = 16 + 12 + 16 + 16 = 60g$

moles of solute = $\frac {4 \times 1000}{60}$

W $\quad$ Weight of solute (kg)

S - W Solvent

molality = $\frac {\text{moles of solute}}{\text {Wt of solvent}}$

$0.25=\frac{w \times 1000}{60(2.5-\omega)}$

$\frac{W}{25-W}=\frac{0.25 \times 6 \phi}{100 \phi}=1.5 \times 10^{-2}=\frac{.015}{1}$

$\frac{w}{25-w+w}=\frac{0.015}{1+0.015} \Rightarrow h=\frac{0.015}{1015} \times 2.5$

$=0.037 \mathrm{Kg}$

$=37g$

a. molality

b. molarity

c. mole fraction

20% mass/mass $\quad$ f = 1.202g/ml

20g $\quad$ K I(39.1 126q) = 1660g/mol

80g $\quad$ solvent water

$^nKI = \frac{20}{166}$ = 0.12moles

molality = $\frac {\text{moles of solvent}}{\text{water of solvent}}$

$= \frac {0.12}{80} \times 1000 = \frac {120}{80} = 1.5m$

molarity = $\frac {\text{moles of solute}}{\text{volume of solution}}$

= $\frac {0.12}{83.19} \times 1000 = 1.44$

V = $\frac {100g}{1.202g/ml} = 83.19ml$

$^XKI = \frac {4^nKI}{^nKI + ^nH_2O} = \frac {0.12}{0.12+ \frac {80}{18}} = 0.026$

$300\mathrm{g} \text { of } 25$ $+400 \mathrm{g}\text { of } 40$

$25 \times 300 / 100=75 \mathrm{g}$

$400 \times 40 / 100=160 \mathrm{g}$

Mass percentage $=\frac{\text { mass of the component }}{\text { Total mass }}$

$=\frac{235}{700} \times 100=33.6$

1. molality = $\frac {\text {moles of solute}}{\text{Wt of solvent}}$

= 3.59/.2 = 17.95m

3.95mls = $\frac {222.6g}{62g/mol}$ ethylene glycol

200g of water

$\rho = 1.072g/ml$

V = $\frac {422.6g}{1.072g/ml}$

= 394.2ml

molarity =$\frac {\text{moles of solute}}{\text {volume of solution}}$

= $\frac{3.59}{3942}$

= 9.11 M