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What will be the temperature at which a solution of $20 \mathrm{~g}$ of $\mathrm{C}_4 \mathrm{H}_{10}$ in $300 \mathrm{~g}$ of $\mathrm{C}_6 \mathrm{H}_6$ freezes, the molal freezing point depression constant of $\mathrm{C}_6 \mathrm{H}_6$ is $5.12 . \mathrm{C_6 H_6}$ freezes at $8.2^{\circ} \mathrm{C}$ .
Option: 1
$10^{\circ} \mathrm{C}$

Option: 2
$3^{\circ} \mathrm{C}$

Option: 3
$5^{\circ} \mathrm{C}$

Option: 4
$8^{\circ} \mathrm{C}$

Answers (1)

best_answer

$\mathrm{ T_f^0=8.2 \mathrm{C} }$
$\mathrm{ K_f=5.12^{\circ} \mathrm{C} / \mathrm{m} \Rightarrow m=300 \mathrm{~g} }$
msolute =20
moler mass of solute = 58
$\mathrm{ \left(\mathrm{C}_4 \mathrm{H}_{10}\right) }$
$\mathrm{ i=1}$

$\mathrm{ \quad \Delta T_f=i \mathrm{~K}+m }$

$\mathrm{ T_f^0-T_f^1=1 \times 5.12 \times \frac{20 / 58}{300 / 1000} }$
$\mathrm{ T_f^0-T_f^1=1 \times 5.12 \times \frac{20}{58} \times \frac{1000}{300} \\ }$

$\mathrm{ T_f^0-T_f^1=5.88 }$
$\mathrm{ 8.2-T_f^1=5.88 }$
$\mathrm{T_f^1=2.32^{\circ} \mathrm{C} }$
$\mathrm{ T_f^1 \approx 3^{\circ} \mathrm{C} }$ .

Posted by

HARSH KANKARIA

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