Question

the relative lowering of vapour pressure of a solution containing 34.2g of sucrose in 900g of water is
a.1/50 b.1/51 c.1/500 d5.1/501
​

Answer 1

$\bigstar$ Given

A solution containing 34.2 g of sucrose in 900 g of water

$\bigstar$ To Find

The relative lowering of vapour pressure

$\bigstar$ Solution

We know that

Molar mass of sucrose = 342 g/mol

Given that

A solution containing 34.2 g of sucrose in 900 g of water

Hence

Number of moles of sucrose (n)

$\sf \longrightarrow n= \dfrac{34.2}{342} \ mol$

$\sf \longrightarrow n= 0.1 \ mol$

Therefore

Number of moles of sucrose = 0.1 mol

We know that

Molar mass of water = 18 g/mol

Given that

A solution containing 34.2 g of sucrose in 900 g of water

Hence

Number of moles of water (n')

$\sf \longrightarrow n'= \dfrac{900}{18} \ mol$

$\sf \longrightarrow n'= 50 \ mol$

According to the question

We are asked to find the relative lowering of vapour pressure

Therefore

The relative lowering of the vapour pressure will be equal to the mole fraction of sucrose

Hence

$\red{\sf \longrightarrow M=\dfrac{n}{n+n'}}$

Here

• M = relative lowering of vapour pressure
• n = number of moles of sucrose
• n' = number of moles of water

Substituting the values

We get

$\sf \longrightarrow M=\dfrac{0.1}{0.1+50}$

$\sf \longrightarrow M=\dfrac{0.1}{50.1}$

$\sf \longrightarrow M=\dfrac{0.1}{50.1} \times \dfrac{10}{10}$

Therefore

$\pink{\sf \longrightarrow M=\dfrac{1}{501}}$

So

The relative lowering of vapour pressure = 1/501

Hence

$\checkmark$ Option (d) is correct