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Section: Chemistry

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Question : 8 of 29

Marks: +1, -0

1.95 g of non-volatile and non-electrolyte solute dissolved in 100 g of benzene lowered the freezing point of it by 0.64 K . The molar mass of the solute (in g

\mathrm{mol}^{-1}) \;\; (K_f (\mathrm{C}_6\mathrm{H}_6) =5.12 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1})

[AP EAPCET 22 May 2025 S2]

$240$
$156$
$165$
$265$

Solution:

Depression in freezing point is given by

\Delta T_f\;=K_f \cdot m

\Rightarrow \;\; m\;=\;\frac{\Delta T_f}{K_f}=\;\frac{0.64}{5.12}=0.125 \text{ mol/kg}

Mole of solute

=

molality × mass of solvent

\;=0.125 \times \;\frac{100}{1000}=0.0125 \text{ mol}

\text{ Molar mass }\;\;=\;\frac{\text{ Mass of solute }}{\text{ Moles of solute }}=\;\frac{1.95}{0.0125}=156 \text{ g/mol}

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