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When 3.00 g of a substance

Question:

When $3.00 \mathrm{~g}$ of a substance ' $\mathrm{X}$ ' is dissolved in $100 \mathrm{~g}$ of $\mathrm{CCl}_{4}$, it raises the boiling point by $0.60 \mathrm{~K}$. The molar mass of the substance ' $\mathrm{X}^{\prime}$ is g $\mathrm{mol}^{-1}$. (Nearest integer).

$\left[\right.$ Given $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{CCl}_{4}$ is $\left.5.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$

Solution:

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times$ molality

$0.60=5 \times\left(\frac{3 / \mathrm{M}}{100 / 100}\right)$

$\mathrm{M}=250$

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