Verified Solution • 2025–26 Session moderate

QJEEM22Ch6c10: JEE Main 2022 Chemistry - Class 12 JEEM22Ch6c10 Solution: of solute A was dissolved in of ethanol and freezing point of…

1.80 g1.80 \mathrm{~g} of solute A was dissolved in 62.5 cm362.5 \mathrm{~cm}^{3} of ethanol and freezing point of the solution was found to be 155.1 K155.1 \mathrm{~K} . The molar mass of solute A is ________ g mol−1\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]

Step-by-Step Verified Solution

Mass of solvent =dxv=0.8×62.5=50=d x v=0.8 \times 62.5=50 gram

ΔTf=kf×m\Delta T_{f}=k_{f} \times m

amp;0.9=2[1.8×1000MSolute ×50]amp;MSolute =(2×1.8×10000.9×50)=80\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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