MHT CET · Chemistry · Solutions
What is the molar mass of solute if a solution is prepared by dissolving 4.5 gram solute in \(3 \mathrm{dm}^3\) water having osmotic pressure \(0.25 \mathrm{~atm}\) at \(300 \mathrm{~K}\)?
\(\left(\mathrm{R}=0.0821 \mathrm{dm}^3 \mathrm{~atm} \mathrm{~K} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(148 \mathrm{~g} \mathrm{~mol}^{-1}\)
- B \(160 \mathrm{~g} \mathrm{~mol}^{-1}\)
- C \(172 \mathrm{~g} \mathrm{~mol}^{-1}\)
- D \(136 \mathrm{~g} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(A) \(148 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\pi=\mathrm{CRT}\)
\(0.25=\frac{4.5}{\mathrm{M}_{\mathrm{B}}} \times \frac{1000}{3000} \times 0.0821 \times 300\)
\(\mathrm{M}_{\mathrm{B}}=147.78 \mathrm{~g} / \mathrm{mol}\)
\(0.25=\frac{4.5}{\mathrm{M}_{\mathrm{B}}} \times \frac{1000}{3000} \times 0.0821 \times 300\)
\(\mathrm{M}_{\mathrm{B}}=147.78 \mathrm{~g} / \mathrm{mol}\)
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