MHT CET · Chemistry · Solutions
Calculate molar mass of a solute at 300 K if 400 mg of it is dissolved in 300 mL of water exerts osmotic pressure of 0.2 atm .
\(\left(\mathrm{R}=0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(90 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(120 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(164 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(180 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(C) \(164 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M}_2 =\frac{\mathrm{W}_2 \mathrm{RT}}{\pi \mathrm{V}} \)
\( =\frac{0.4 \mathrm{~g} \times 0.0821 \mathrm{dm}^3 \mathrm{atmK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.2 \mathrm{~atm} \times 0.3 \mathrm{dm}^3} \)
\( =164.2 \mathrm{~g} \mathrm{~mol}^{-1} \approx 164 \mathrm{~g} \mathrm{~mol}^{-1}\)
\( =\frac{0.4 \mathrm{~g} \times 0.0821 \mathrm{dm}^3 \mathrm{atmK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.2 \mathrm{~atm} \times 0.3 \mathrm{dm}^3} \)
\( =164.2 \mathrm{~g} \mathrm{~mol}^{-1} \approx 164 \mathrm{~g} \mathrm{~mol}^{-1}\)
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