Question

0 .15 g of a substance dissolved in 15 g of a solvent higher by 0.216$^\circ$C than that of the pure solvent. Find out the molecular weight of the substance. $(K_b\, for\, solvent \, is\, 2.16 ^\circ C)$

• A. 1.01
• B. 10.1
• C. 100
• D. 10

Answer 1

The Correct Option is C

To find the molecular weight (molar mass) of the substance using the given data, we apply the formula for boiling point elevation:

\(\Delta T_b = i \cdot K_b \cdot \frac{m}{M}\)

Where:

• \(\Delta T_b\) is the boiling point elevation.
• i is the van't Hoff factor (for non-electrolytes, i = 1).
• K_b\) is the ebullioscopic constant (2.16 ^{\circ}C\)).
• m is the mass of the solute (0.15 g).
• M is the molar mass of the solute (what we're trying to find).
• The mass of the solvent in kg is needed (15 g = 0.015 kg).

Given that the elevation in boiling point is 0.216^{\circ}C, rearrange the formula to solve for M:

M = \frac{i \cdot K_b \cdot m}{\Delta T_b \cdot \text{mass of solvent in kg}}

Substitute the values:

M = \frac{1 \cdot 2.16 \cdot 0.15}{0.216 \cdot 0.015}

M = \frac{0.324}{0.00324}

M = 100

Thus, the molecular weight of the substance is 100, which matches the given option.