SATHEE: Solutions Question 118

Solutions Question 118

Question: The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at $25^{o} C$ is 23.8 mm Hg [RPET 1999]

Options:

A) 1.25 mm Hg

B) 0.125 mm Hg

C) 1.15 mm Hg

D) 00.12 mm Hg

Show Answer

Answer:

Correct Answer: B

Solution:

Given molecular mass of sucrose = 342 Moles of sucrose $= \frac{100}{342} = 0.292$ mole Moles of water $N = \frac{1000}{18} = 55.5$ moles and Vapour pressure of pure water $P^{0} = 23.8$ mm Hg According to Raoul’s law $\frac{Δ P}{P^{0}} = \frac{n}{n + N} \Rightarrow \frac{Δ P}{23.8} = \frac{0.292}{0.292 + 55.5}$

$Δ P = \frac{23.8 \times 0.292}{55.792} = 0.125$ mm Hg.

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Solutions Question 118

Question: The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25oC is 23.8 mm Hg [RPET 1999]

Options:

Answer:

Solution:

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Question: The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at $25^{o} C$ is 23.8 mm Hg [RPET 1999]