Question

2 g nonelectrolyte solute when dissolved in 50 g water increases its boiling point by 0.13 K.
Calculate the molar mass of solute if molal elevation constant of water is 0.52 K kg mol\(^{-1}\)

• A. \(140\) g/mol
• B. \(150\) g/mol
• C. \(160\) g/mol
• D. \(170\) g/mol

Hint:

Use the elevation formula with mass of solvent in kilograms, then rearrange it for molar mass.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When a non-volatile solute dissolves in a solvent, the boiling point rises in proportion to the molality of the solution.

Step 2: Key Formula or Approach:
\[ \Delta T_b = K_b \times m, \qquad m = \frac{w_2 \times 1000}{M_2 \times w_1\ (\text{in g})} \]

Step 3: Detailed Explanation:
Given \(w_2 = 2\) g, \(w_1 = 50\) g, \(\Delta T_b = 0.13\) K and \(K_b = 0.52\text{ K kg mol}^{-1}\).
\[ M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} = \frac{0.52 \times 2 \times 1000}{0.13 \times 50} \]
\[ M_2 = \frac{1040}{6.5} = 160\text{ g/mol} \]

Step 4: Why the other options are wrong.
Putting 140, 150 or 170 back into the formula gives \(\Delta T_b\) of 0.149 K, 0.139 K and 0.122 K. None equals 0.13 K. Only 160 g/mol does.

Final Answer:
The molar mass of the solute is 160 g/mol, option (C). \[ \boxed{160\text{ g/mol}} \]