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The depression in freezing point observed for a formic acid solution of concentration $0.5 \mathrm{~mL} \mathrm{~L}^{-1}$ is $0.0405^{\circ} \mathrm{C}$ . Density of formic acid is $1.05 \mathrm{~g} \mathrm{~mL}^{-1}$ . The Van't Hoff factor of the formic acid solution is nearly :
(Given for water $\mathrm{k}_{\mathrm{f}}=1.86 \; \mathrm{k}\; \mathrm{kg}\; \mathrm{mol}^{-1}$ )
Option: 1
0.8

Option: 2
1.1

Option: 3
1.9

Option: 4
2.4

Answers (1)

best_answer

We know, $\mathrm{\Delta T=ik_{f}m......(1)}$

given, $\mathrm{k_{f}}=1.86\; \mathrm{k} \; \mathrm{Kg}\; \mathrm{mol}^{-1}\; (\mathrm{solvent=water})$

$\mathrm{molality(m)=\frac{mole\; of \; solute}{Solvent(kg)}}$

Now, concentration of formic acid $\mathrm{[HCOOH]=0.5\; mL\; L^{-1}}$ means $\mathrm{0.5\; mL\;of\; [HCOOH]}$ is present in 1 litre $\mathrm{H_{2}O}$

So, Mass $\mathrm{HCOOH=0.5\; ml \times Density=0.5\times1.05 \; g=0.525\; g}$

So, 0.525 g of $\mathrm{HCOOH}$ is present in 1 litre of water

then $\mathrm{Molality=\frac{Mass \; of \; HCOOH}{Molar\; mass\; of\; HCOOH \times Solvent(kg)}}$

$\mathrm{m=\frac{0.525}{4.6 \times 1}\; \; \; \; \; \; \; [\because 1 L, H_{2}O=1 kg \; of\; H_{2}O]}$

Put all value in (1)

$\mathrm{\Delta T_{f}=i k_{f} m=i \times 1.86 \times \frac{0.525}{46 \times 1}}$

$\mathrm{0.0405 =i \times 1.86 \times \frac{0.525}{46}}$

$\mathrm{i = 1.9 (Van's \; Hoff\; factor)}$

Hence, the correct answer is Option (3).

Posted by

Ajit Kumar Dubey

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