MHT CET · Chemistry · Solutions
The solution containing \(3 \mathrm{~g}\) urea (molar mass 60 ) per \(\mathrm{dm}^3\) of water and another solution containing \(4.5 \mathrm{~g}\) of solute A per \(\mathrm{dm}^3\) boils at same temperature, then what is molar mass of A?
- A \(54 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(180 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(120 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(90 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(D) \(90 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \cdot \mathrm{m}\)
\(\text { molality }(\mathrm{m})=\frac{\text { moles of solute }}{\text { mass of water }(\mathrm{kg})}\)
1L \(\mathrm{H}_2 \mathrm{O}=1 \mathrm{dm}^3 \mathrm{H}_2 \mathrm{O}=1 \mathrm{~kg} \mathrm{H}_2 \mathrm{O}\)
Density of water \(=1 \mathrm{~kg} / \mathrm{L}\)
Boiling points are same for two solutions,
\(\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}_1}=\Delta \mathrm{T}_{\mathrm{b}_2} \\
& \mathrm{~m}_1=\mathrm{m}_2
\end{aligned}\)
\(\begin{aligned}
& \frac{3 / 60}{1}=\frac{4.5 / \mathrm{M}}{1} \\
& \mathrm{M}=\frac{60 \times 4.5}{3} \\
& =90 \mathrm{~g} / \mathrm{mol}
\end{aligned}\)
\(\text { molality }(\mathrm{m})=\frac{\text { moles of solute }}{\text { mass of water }(\mathrm{kg})}\)
1L \(\mathrm{H}_2 \mathrm{O}=1 \mathrm{dm}^3 \mathrm{H}_2 \mathrm{O}=1 \mathrm{~kg} \mathrm{H}_2 \mathrm{O}\)
Density of water \(=1 \mathrm{~kg} / \mathrm{L}\)
Boiling points are same for two solutions,
\(\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}_1}=\Delta \mathrm{T}_{\mathrm{b}_2} \\
& \mathrm{~m}_1=\mathrm{m}_2
\end{aligned}\)
\(\begin{aligned}
& \frac{3 / 60}{1}=\frac{4.5 / \mathrm{M}}{1} \\
& \mathrm{M}=\frac{60 \times 4.5}{3} \\
& =90 \mathrm{~g} / \mathrm{mol}
\end{aligned}\)
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