Question

A solution of \(50\)g of solute 'X' is dissolved in \(150\) g of 'Y' solvent boils at \(357.27\) K. What is the molar mass of solute?
(Given \(K_b\) and boiling point pure solvent is \(2.77\) K kg/mol and \(350.06\) K respectively.)

• A. \(132\) g/mol
• B. \(128\) g/mol
• C. \(150\) g/mol
• D. \(15\) g/mol

Hint:

Use delta Tb = Kb x m, then find moles of solute and the molar mass.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Adding a non-volatile solute raises the boiling point of the solvent. The rise is proportional to the molality of the solution.

Step 2: Key Formula:
\[ \Delta T_b = K_b \cdot m = K_b \cdot \dfrac{w_2 \times 1000}{M_2 \times w_1} \]

Step 3: Find the elevation:
\(\Delta T_b = 357.27 - 350.06 = 7.21\) K.

Step 4: Solve for the molar mass:
\[ M_2 = \dfrac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} = \dfrac{2.77 \times 50 \times 1000}{7.21 \times 150} \]
\[ M_2 = \dfrac{138500}{1081.5} = 128.06 \approx 128 \text{ g/mol} \]

Step 5: Why the other options are wrong.
132 and 150 are close guesses without the correct arithmetic. 15 g/mol is off by a factor of ten. Only 128 g/mol follows from the formula.

Final Answer:
The molar mass of the solute is 128 g/mol. \[ \boxed{\text{(B) }128\ \text{g/mol}} \]