MHT CET · Chemistry · Solutions
A solution of nonvolatile solute is obtained by dissolving 0.8 g in \(0.3 \mathrm{dm}^3\) water has osmotic pressure 0.2 atm at 300 K . Calculate the molar mass of solute.
\(\left[\mathrm{R}=0.082 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\)
- A \(300 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(340 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(328 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(352 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(328 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M}_2 =\frac{\mathrm{W}_2 \mathrm{RT}}{\pi \mathrm{V}}\)
\(=\frac{0.8 \mathrm{~g} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.2 \mathrm{~atm} \times 0.3 \mathrm{dm}^3}\)
\(=328 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(=\frac{0.8 \mathrm{~g} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.2 \mathrm{~atm} \times 0.3 \mathrm{dm}^3}\)
\(=328 \mathrm{~g} \mathrm{~mol}^{-1}\)
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