Question

A solution of 50 g of non-volatile solute in 150 g of ethyl acetate shows a boiling point of 84.27 deg * C If boiling point and K b 1 for ethyl acetate are 77.27 deg * C and 2.73 K kgmo * l ^ - 1 respectively then find molar mass of solute. 28. Mo 2.7

• A. 120gmo * l ^ - 1
• B. 130 gmol^ -1
• C. 140 g mol
• D. 160 gmgl^ -1

Answer 1

Answer: (B)

130 g mol$^{-1}$

Answer 2

Solution:

1. Calculate Boiling Point Elevation:

   $$\Delta T = T_{\text{solution}} - T_{\text{pure solvent}} = 84.27^\circ C - 77.27^\circ C = 7.00^\circ C$$

2. Determine the Molality (m):

   Using the formula, $\Delta T = K_b \times m$:

   $$m = \frac{\Delta T}{K_b} = \frac{7.00}{2.73} \approx 2.562\, \text{mol/kg}$$

3. Calculate Moles of Solute:

   Given the mass of solvent = 150 g = 0.150 kg,

   $$\text{Moles of solute} = m \times (\text{kg of solvent}) = 2.562 \times 0.150 \approx 0.3843\, \text{moles}$$

4. Determine Molar Mass:

   $$\text{Molar Mass} = \frac{\text{Mass of solute}}{\text{Moles of solute}} = \frac{50\, \text{g}}{0.3843\, \text{moles}} \approx 130\, \text{g/mol}$$