KCET · Chemistry · Solutions
The mass of a non-volatile solute of molar mass \(40 \mathrm{~g} \mathrm{~mol} \mathrm{~m}^{-1}\) that should be dissolved in \(114 \mathrm{~g} \mathrm{of}\) octane to lower its vapour pressure by \(20 \%\) is
- A \(11.4 \mathrm{~g}\)
- B \(9.8 \mathrm{~g}\)
- C \(12.8 \mathrm{~g}\)
- D \(10 \mathrm{~g}\)
Answer & Solution
Correct Answer
(D) \(10 \mathrm{~g}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{p}^{0}=100, \quad \mathrm{p}=100-20=80\)
\(\mathrm{m}=40 \quad \mathrm{M}\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)=114\)
\(\mathrm{w}=? \quad \mathrm{~W}=114\)
\(\because \quad \frac{\mathrm{p}^{\circ}-\mathrm{p}}{\mathrm{p}}=\frac{\mathrm{w}}{\mathrm{m}} \times \frac{\mathrm{M}}{\mathrm{W}}\)
\(\therefore \quad \frac{100-80}{80}=\frac{w}{40} \times \frac{114}{114}\)
or \(\quad \frac{20}{80}=\frac{w}{40}\)
or \(\quad \mathrm{w}=\frac{20 \times 40}{80}=10 \mathrm{~g}\)
\(\mathrm{m}=40 \quad \mathrm{M}\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)=114\)
\(\mathrm{w}=? \quad \mathrm{~W}=114\)
\(\because \quad \frac{\mathrm{p}^{\circ}-\mathrm{p}}{\mathrm{p}}=\frac{\mathrm{w}}{\mathrm{m}} \times \frac{\mathrm{M}}{\mathrm{W}}\)
\(\therefore \quad \frac{100-80}{80}=\frac{w}{40} \times \frac{114}{114}\)
or \(\quad \frac{20}{80}=\frac{w}{40}\)
or \(\quad \mathrm{w}=\frac{20 \times 40}{80}=10 \mathrm{~g}\)
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