The vapour pressure of water at 20 ° C is 17 .5 mm of Hg . If 18 g of glucose C 6 H 12 O 6 is added to 178 . 2 g of water at 20 ∘ C , what will be the vapour pressure of the resulting solution?

Moles of C 6 H 12 O 6 = 18 180 = 0 .1 Moles of H 2 O = 178 .2 18 9 .9 According to Raoult's law P ° - P s P ° = X solute 17 .5 - P s 17 .5 = 0 .1 10 so, P s = 17 . 325 mm of Hg

Chemistry - Liquid Solution Question with Solution | TestHub

ChemistryLiquid SolutionRaoult's LawEasy2 minPYQ_2020

ChemistryEasysingle choice

The vapour pressure of water at $20 ° C$ is $17 .5 mm of Hg .$ If $18 g$ of glucose $(C_{6} H_{12} O_{6})$ is added to $178.2 g$ of water at $20^{\circ} C$ , what will be the vapour pressure of the resulting solution?

Options:

Answer:

Solution:

Moles of $C_{6} H_{12} O_{6} = \frac{18}{180} = 0 .1$

Moles of $H_{2} O = \frac{178 .2}{18} 9 .9$

According to Raoult's law

$\frac{P^{°} - P_{s}}{P^{°}} = X_{solute}$

$\frac{17 .5 - P_{s}}{17 .5} = \frac{0 .1}{10}$

so, $P_{s} = 17.325 mm of Hg$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Liquid SolutionSubtopic:Raoult's Law

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...