JEE Main & Advanced Numericals

1. Atomic radius of boron: $$V={\displaystyle \frac{m}{d}}$$

$$V={\displaystyle \frac{10.81\mathtt{\text{ g/mol}}}{2.34{\mathtt{\text{ g/cm}}}^3}}$$

$$V=4.62{\mathtt{\text{ cm}}}^3/\mathtt{\text{mol}}$$

$$r={\left({\displaystyle \frac{3V}{4\pi N_A}}\right)}^{\frac{1}{3}}$$

$$r={\left({\displaystyle \frac{3\times 4.62{\mathtt{\text{ cm}}}^3/\mathtt{\text{mol}}}{4\pi \times 6.022\times {10}^{23}{\mathtt{\text{ mol}}}^{-1}}}\right)}^{\frac{1}{3}}$$

$$r=0.88\times {10}^{-8}\mathtt{\text{ cm}}$$

Atomic radius of boron = 88 pm

2. Ionization energy of gallium:

$$I{E}_{\text{Ga}}=I{E}_{\text{Al}}+\mathrm{\Delta }IE$$

$$I{E}_{\text{Ga}}=577\mathtt{\text{ kJ/mol}}+165\mathtt{\text{ kJ/mol}}$$

Ionization energy of gallium = 742 kJ/mol

3. Electronegativity of indium:

$${\chi }_{\text{In}}={\chi }_{\text{Tl}}-\mathrm{\Delta }\chi$$

$${\chi }_{\text{In}}=1.62-0.45$$

Electronegativity of indium = 1.17

4. Melting point of thallium: The melting points of aluminum, indium, and thallium follow a linear trend. Let’s assume the linear equation is:

$$T_{\text{m}}=mx+b$$

where (T_\text{m}) is the melting point, (x) is the atomic number, and (m) and (b) are constants. Using the given data, we can create two equations:

$$\begin{array}{}\text{(1)}& 660=m(13)+b\end{array}$$

$$\begin{array}{}\text{(2)}& 156=m(49)+b\end{array}$$

Subtracting equation (1) from equation (2), we get:

$$-504=m(-36)$$

$$m=14\mathtt{\text{ °C}}$$

Substituting (m) back into equation (1), we get:

$$660=14(13)+b$$

$$b=482\mathtt{\text{ °C}}$$

Now we can use the equation to calculate the melting point of thallium (atomic number 81):

$$T_{\text{m}}=14(81)+482$$

$$T_{\text{m}}=1694\mathtt{\text{ °C}}$$

Melting point of thallium = 1694 °C

5. Boiling point of gallium: The boiling points of aluminum, indium, and gallium follow a linear trend. Let’s assume the linear equation is:

$$T_{\text{b}}=mx+b$$

where (T_\text{b}) is the boiling point, (x) is the atomic number, and (m) and (b) are constants.

Using the given data, we can create two equations:

$$\begin{array}{}\text{(1)}& 2467=m(13)+b\end{array}$$ $$\begin{array}{}\text{(2)}& 2072=m(49)+b\end{array}$$

Subtracting equation (1) from equation (2), we get:

$$-395=m(-36)$$

$$m=11\mathtt{\text{ °C}}$$

Substituting (m) back into equation (1), we get:

$$2467=11(13)+b$$

$$b=2224\mathtt{\text{ °C}}$$

Now we can use the equation to calculate the boiling point of gallium (atomic number 31):

$$T_{\text{b}}=11(31)+2224$$

$$T_{\text{b}}=2575\mathtt{\text{ °C}}$$

Boiling point of gallium = 2575 °C

6. Standard reduction potential of the Al3+/Al couple: Given, (E^\circ_{Ga3+/Ga} = -0.53 \texttt{V}) and (\Delta E^\circ = 0.29 \texttt{V}).

$$E_{Al3+/Al}^{\circ }=E_{Ga3+/Ga}^{\circ }-\mathrm{\Delta }{E}^{\circ }$$

$$E_{Al3+/Al}^{\circ }=-0.53\mathtt{\text{V}}-0.29\mathtt{\text{V}}$$

Standard reduction potential of the Al3+/Al couple = -0.82 V

7. Solubility product constant (Ksp) of In(OH)3: Given, (K_\text{sp,Al(OH)3} = 1.9 \times 10^{-31}) and (\Delta \log K_\text{sp} = 1.5 \times 10^{-31}).

$$\log K_{\text{sp,In(OH)3}}=\log K_{\text{sp,Al(OH)3}}+\mathrm{\Delta }\log K_{\text{sp}}$$

$$\log K_{\text{sp,In(OH)3}}=-30.72+1.5\times {10}^{-31}$$

$$K_{\text{sp,In(OH)3}}=1.9\times {10}^{-31}\times {10}^{1.5\times {10}^{-31}}$$

Solubility product constant of In(OH)3 = 1.98 × 10^⁻³¹

CBSE Board Exam Numericals

1. Atomic mass of boron: The average atomic mass of boron is: $$M_{\text{avg}}=0.199\times 10.01\mathtt{\text{ g/mol}}+0.801\times 11.01\mathtt{\text{ g/mol}}$$ $$M_{\text{avg}}=10.81\mathtt{\text{ g/mol}}$$

Atomic mass of boron = 10.81 g/mol

2. Mass of aluminum produced:

First, calculate the number of moles of Al2O3:

$$n_{\text{Al2O3}}=\frac{100\mathtt{\text{ g}}}{101.96\mathtt{\text{ g/mol}}}=0.98\mathtt{\text{ mol}}$$

Assuming 100% current efficiency, the number of moles of Al produced would be:

$$n_{\text{Al}}=2\times n_{\text{Al2O3}}=1.96\mathtt{\text{ mol}}$$

However, since the current efficiency is 85%, the actual number of moles of Al produced is: $$n_{\text{Al,actual}}=0.85\times 1.96\mathtt{\text{ mol}}=1.67\mathtt{\text{ mol}}$$

Finally, calculate the mass of aluminum produced: $$m_{\text{Al}}=n_{\text{Al,actual}}\times M_{\text{Al}}$$ $$m_{\text{Al}}=1.67\mathtt{\text{ mol}}\times 26.98\mathtt{\text{ g/mol}}=45.2\mathtt{\text{ g}}$$

Mass of aluminum produced = 45.2 g

3. Volume of 0.1 M HCl required:

First, calculate the number of moles of In2O3: $$n_{\text{In2O3}}=\frac{1.0\mathtt{\text{ g}}}{277.64\mathtt{\text{ g/mol}}}=0.0036\mathtt{\text{ mol}}$$

The balanced chemical equation shows that 6 moles of HCl are required for every mole of In2O3. Therefore, the number of moles of HCl required is:

$$n_{\text{HCl}}=6\times n_{\text{In2O3}}=0.022\mathtt{\text{ mol}}$$

Finally, calculate the volume of 0.1 M HCl required: $$V_{\text{HCl}}=\frac{n_{\text{HCl}}}{M_{\text{HCl}}}=\frac{0.022\mathtt{\text{ mol}}}{0.1\mathtt{\text{ mol/L}}}=0.22\mathtt{\text{ L}}$$

Volume of 0.1 M HCl required = 0.22 L

4. Percentage of indium in the alloy:

First, calculate the number of moles of InCl3 produced: $$n_{\text{InCl3}}=\frac{1.34\mathtt{\text{ g}}}{221.18\mathtt{\text{ g/mol}}}=0.00606\mathtt{\text{ mol}}$$

The balanced chemical equation shows that 1 mole of InCl3 is produced for every mole