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Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at \(100^{\circ} \mathrm{C}\) is
(Vapour pressure of pure water at \(100^{\circ} \mathrm{C}=760\) torr)

A 76.0 torr
B 752.0 torr
C 7.6 torr
D 3207.6 torr

Verified Solution

Answer & Solution

Correct Answer

(B) 752.0 torr

Step-by-step Solution

Detailed explanation

Relative lowering of vapour pressure is equal to mole fraction of glucose.
\(\frac{p_0-p_S}{p_0}=\chi_{\text {Glucose }}\) ...(i)
Now, number of moles of glucose \(=\frac{18}{180}=0.1\)
number of moles of water \(=\frac{178.2}{18}=9.9\)
Mole fraction of glucose

\(=\frac{0.1}{0.1+9.9}=0.01\)
Substituting the value of mole fraction in Eq. (i)
\(\frac{760-p}{760}=0.01\)
\(p=752.4\) torr
\(\approx 752.0\) torr

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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Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at \(100^{\circ} \mathrm{C}\) is(Vapour pressure of pure water at \(100^{\circ} \mathrm{C}=760\) torr)

Step-by-step Solution

Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at \(100^{\circ} \mathrm{C}\) is
(Vapour pressure of pure water at \(100^{\circ} \mathrm{C}=760\) torr)