MHT CET · Chemistry · Solutions
A solution of \(8 \mathrm{~g}\) of certain organic compound in \(2 \mathrm{dm}^3\) water develops osmotic pressure \(0.6 \mathrm{~atm}\) at \(300 \mathrm{~K}\). Calculate the molar mass of compound. \(\left[\mathrm{R}=0.082 \mathrm{~atm} \mathrm{dm}^3 \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\)
- A \(148 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(164 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(172 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(180 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(164 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\pi =\frac{\mathrm{W}_2 \mathrm{RT}}{\mathrm{M}_2 \mathrm{~V}} \)
\( \therefore \mathrm{M}_2 =\frac{\mathrm{W}_2 \mathrm{RT}}{\pi \mathrm{V}} \)
\( \therefore \mathrm{M}_2 =\frac{8 \mathrm{~g} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.6 \mathrm{~atm} \times 2 \mathrm{dm}^3} \)
\( =164 \mathrm{~g} \mathrm{~mol}^{-1}\)
\( \therefore \mathrm{M}_2 =\frac{\mathrm{W}_2 \mathrm{RT}}{\pi \mathrm{V}} \)
\( \therefore \mathrm{M}_2 =\frac{8 \mathrm{~g} \times 0.082 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.6 \mathrm{~atm} \times 2 \mathrm{dm}^3} \)
\( =164 \mathrm{~g} \mathrm{~mol}^{-1}\)
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