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Liquid Solution

Section: Chemistry

Question : 1 of 29

Marks: +1, -0

An aqueous solution containing 0.2 g of a non volatile solute ' A ' in 21.5 g of water freezes at 272.814 K . If the freezing point of water is 273.16 K , the molar mass (in

\mathrm{g} \mathrm{mol}^{-1}

) of solute

A

[ K_f (\mathrm{H}_2\mathrm{O}) = 1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} ]

[AP EAPCET 24 May 2025 S1]

Validate

Solution:

Step 1: Find the decrease in freezing point

The difference between the normal freezing point of water and the solution's freezing point is called the depression in freezing point. The calculation is:

\; \Delta T_f = T_f^0 - T_f = 273.16 - 272.814

\; \Delta T_f = 0.346 \mathrm{K}

Step 2: Use the freezing point depression formula

The formula is:

\Delta T_f = K_f \cdot m

where

m

is molality.

m = \; \frac{\Delta T_f}{K_f} = \; \frac{0.346}{1.86} = 0.186 \mathrm{mol/kg}

Step 3: Calculate the moles of solute

Molality tells us moles of solute in 1 kg of solvent (water). Here, amount of water is

21.5 \mathrm{g} = 0.0215 \mathrm{kg}

Moles of solute

= 0.186 \times 0.0215 = 0.00400 \mathrm{mol}

Step 4: Calculate the molar mass

Molar mass

= \frac{\text{mass of solute}}{\text{moles of solute}}

Molar mass

= \frac{0.2}{0.00400} = 50 \mathrm{g/mol}

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