MHT CET · Chemistry · Solutions

9 gram anhydrous oxalic acid (mol. Wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is $P_{1}^{o} c$ the vapour pressure of solution is

A $0.99 P_{1}^{o}$
B $0.1 P_{1}^{o}$
C $0.9 1 P_{1}^{o}$
D $1.1 P_{1}^{o}$

Verified Solution

Answer & Solution

Correct Answer

(A) $0.99 P_{1}^{o}$

Step-by-step Solution

Detailed explanation

The total vapour pressure of a solution in this case only depends on vapour pressure of water as anhydrous oxalic acid is a non-volatile compound.

∴

Vapour pressure of solution = vapour pressure of
Water

(P_{w})

According to Raoult’s law
Number of moles of oxalic acid =

\frac{9}{90} = 0.1

moles

∴ x_{w} = \frac{9.9}{9.9 + 0.1} = 0.99

\Rightarrow P_{s} = P_{w} = 0.99 \times P_{1}^{o}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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9 gram anhydrous oxalic acid (mol. Wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is P1oc the vapour pressure of solution is

Step-by-step Solution

9 gram anhydrous oxalic acid (mol. Wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is $P_{1}^{o} c$ the vapour pressure of solution is