Question

When 3.00 g of a substance of molar mass 250 g mol$^{-1}$ is dissolved in 100 g of CCl$_4$, the boiling point of the solvent (CCl$_4$) will be elevated by _ _ _ K. (rounded off to two decimal places)

Hint:

For elevation in boiling point, always use mass of solvent in kg while calculating molality

Answer 1

Correct Answer: 0.6

Step 1: Formula.
\[ \Delta T_b = K_b\, m \]

Step 2: Find the molality.
Moles of solute are $3.00/250 = 0.012$, and in 0.100 kg of solvent the molality is 0.12.

Step 3: Multiply.
\[ \Delta T_b = 5.00 \times 0.12 = 0.60 \text{ K} \]

Step 4: Answer.
\[ \boxed{0.60 \text{ K}} \]