$\mathrm{C}_{6} \mathrm{H}_{6}$ freezes at $5.5^{\circ} \mathrm{C}$. The temperature at which a solution of 10 g of $\mathrm{C}_{4} \mathrm{H}_{10}$ in 200 g of $\mathrm{C}_{6} \mathrm{H}_{6}$ freeze is $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$ (The molal freezing point depression constant of $\mathrm{C}_{6} \mathrm{H}_{6}$ is $5.12^{\circ} \mathrm{C} / \mathrm{m}$)

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Solutions

C_{6} H_{6}

freezes at

5. 5^{\circ} C

. The temperature at which a solution of 10 g of

C_{4} H_{10}

in 200 g of

C_{6} H_{6}

freeze is

____

^{\circ} C

(The molal freezing point depression constant of

C_{6} H_{6}

5.1 2^{\circ} C / m

)

Answer: 1

Solution

Step 1: Calculate molality of the solution.

Molar mass of

C X_{4} H X_{10}

= 58 g/mol

Moles of

C X_{4} H X_{10}

\frac{10}{58} = 0.1724

mol

Mass of benzene = 200 g = 0.2 kg

Molality

m = \frac{0.1724}{0.2} = 0.862

mol/kg

Step 2: Calculate freezing point depression.

Δ T_{f} = K_{f} \times m = 5.12 \times 0.862 = 4.4 1^{\circ}

C

Step 3: Calculate freezing point of solution.

T_{f} = 5.5 - 4.41 = 1.09 \approx 1^{\circ}

C

Answer: 1°C

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Numerical answer

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