JEE MAIN - Chemistry (2022 - 29th July Evening Shift - No. 17)
$$1.80 \mathrm{~g}$$ 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 A is ________ g $$\mathrm{mol}^{-1}$$.
[Given : Freezing point of ethanol is 156.0 K.
Density of ethanol is 0.80 g cm$$-$$3.
Freezing point depression constant of ethanol is 2.00 K kg mol$$-$$1]
Answer
80
Explanation
Mass of solvent $$=d x v=0.8 \times 62.5=50$$ gram
$$\Delta T_{f}=k_{f} \times m$$
$$ \begin{aligned} &0.9=2\left[\frac{1.8 \times 1000}{M_{\text {Solute }} \times 50}\right] \\ &M_{\text {Solute }}=\left(\frac{2 \times 1.8 \times 1000}{0.9 \times 50}\right)=80 \end{aligned} $$
$$\Delta T_{f}=k_{f} \times m$$
$$ \begin{aligned} &0.9=2\left[\frac{1.8 \times 1000}{M_{\text {Solute }} \times 50}\right] \\ &M_{\text {Solute }}=\left(\frac{2 \times 1.8 \times 1000}{0.9 \times 50}\right)=80 \end{aligned} $$
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