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{\Delta P}{P^{0}}=\frac{n}{n+N}\Rightarrow \frac{\Delta P}{23.8}=\frac{0.292}{0.292+55.5} $
$ \Delta P=\frac{23.8\times 0.292}{55.792}=0.125 $ mm Hg.
