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Solutions Part 2

Section: Chemistry

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Question : 93 of 95

Marks: +1, -0

The depression in freezing point observed for a formic acid solution of concentration

0.5 \mathrm{mL} \mathrm{L}^{-1}

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

k_f = 1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1}

[25-Jul-2022-Shift-1]

$0.8$
$1.1$
$1.9$
$2.4$

Solution:

\Delta T_f

of formic acid

=0.0405^{\circ} \mathrm{C}

Concentration

=\frac{0.5 \mathrm{mL}}{\mathrm{L}}

and density

=\frac{1.05 \mathrm{g}}{\mathrm{mL}}

\therefore

Mass of formic acid in solution

=1.05 \times 0.5 \mathrm{g}

=0.525 \mathrm{g}

\therefore

According to Van't Hoff equation,

\Delta T_f = i k_f \cdot m

0.0405 = i \times 1.86 \times \;\frac{0.525}{46 \times 1}

(Assuming mass of

1 \mathrm{L}

water

= \mathrm{kg}

)

i = \frac{0.0405 \times 46}{1.86 \times 0.525} = 1.89 \approx 1.9

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