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: Use elevation in boiling point formula.
\[ \Delta T_b = K_b m \]
where $K_b$ is boiling point constant and $m$ is molality

Step 2: Calculate moles of solute.
\[ \text{Moles of solute} = \frac{3.00}{250} \]
\[ = 0.012 \text{ mol} \]

Step 3: Convert mass of solvent into kg.
\[ 100 \text{ g} = 0.100 \text{ kg} \]

Step 4: Calculate molality.
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
\[ m = \frac{0.012}{0.100} \]
\[ m = 0.12 \text{ mol kg}^{-1} \]

Step 5: Substitute values.
\[ \Delta T_b = 5.00 \times 0.12 \]
\[ \Delta T_b = 0.60 \text{ K} \]

Step 6: Conclusion.
\[ \boxed{0.60 \text{ K}} \]