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  • of solute A was dissolved in <img alt="62.5 \mathrm{~cm}^{3}" src="https://entranceco

1.80 \mathrm{~g} of solute A was dissolved in 62.5 \mathrm{~cm}^{3} of ethanol and freezing point of the solution was found to be 155.1 \mathrm{~K}.. The molar mass of solute \mathrm{A} is ___ \mathrm{g}\: \mathrm{mol}^{-1}.  [Given: Freezing point of ethanol is 156.0 \mathrm{~K}.
Density of ethanol is 0.80 \mathrm{~g} \mathrm{~cm}^{-3}.
Freezing point depression constant of ethanol is 2.00 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ]

Option: 1

80


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\text{wt.of solute }=1.80\, \mathrm{g}}
\mathrm{\text{val.of solvent }=62.5\, \mathrm{ml}}
\mathrm{\text{density of solvent} =0.8 \mathrm{~g} / \mathrm{ml}}
\mathrm{\text{weight of solvent} =d \times v=50 \mathrm{~g}}
\mathrm{\text{Depression constant }= \left ( k_{f} \right )= 2\, k\, kg\, mol^{-1}}
\mathrm{\text{Depression in freezing point }= 0.9\, k}

\mathrm{\Delta t_{f} =k_{f} \times \frac{\text { wt.of solute } \times 1000}{\text { molar mass of solute } \times wt .of \text { solvent }(g)} }
\mathrm{0.9 =\frac{2 \mathrm{k} \, \mathrm{kg} / \mathrm{mol} \times 1.8 \mathrm{~g} \times 1000}{m \times 50 \mathrm{~g}} }
\mathrm{m =\frac{2 \times 1.8 \times 1000}{0.8\times 50}g/mol }
     \mathrm{ =80g/mol }

Hence answer is (80).

Posted by

Shailly goel

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