TS EAMCET · Chemistry · Solutions
The vapour pressure of pure water is \(23 \mathrm{mmHg}\). The vapour pressure of an aqueous solution, which contains 10 mass per cent of solute ' \(A\) ' having molecular weight 50 is
- A \(0.003 \mathrm{~atm}\)
- B \(34.5 \mathrm{~atm}\)
- C \(22 \mathrm{~atm}\)
- D \(0.028 \mathrm{~atm}\)
Answer & Solution
Correct Answer
(D) \(0.028 \mathrm{~atm}\)
Step-by-step Solution
Detailed explanation
Given, Vapour pressure of pure water \(=23 \mathrm{~mm} \mathrm{Hg} 10 \%\) of solute \(A\) Molecular weight of \(A=50\) Mass of \(A=10 \mathrm{~g}\) Mass of water \(=100-10=90 \mathrm{~g}\) Number of moles of \(A=\frac{10}{50}=0.2\) Number of moles of water \(=\frac{90}{18}=5\)…
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