Question

$34.2g$ of cane sugar is dissolved in $180g$ of water. The relative lowering of vapour of pressure will be:
A: $0.0099$
B: $1.1597$
C: $0.840$
D: $0.9901$

Answer 1

Vapour pressure is defined as that pressure which vapours exert over the liquid under equilibrium. When non-volatile solute is added in a volatile solution, non-volatile particles also come over the surface due to which pressure due to volatile particles decreases.
Formula used: mole fraction$= \dfrac{{{\text{moles of substance}}}}{{{\text{total moles}}}}$
Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$

Complete step by step answer:
Vapour pressure is the pressure which vapours exert over the liquid under equilibrium. When non-volatile solute is added in a volatile solution, non-volatile particles also come over the surface due to which pressure due to volatile particles decreases as concentration of volatile particles on the surface decreases. Due to decrease in volatile particles over the surface, pressure due to these particles is also decreased. This decrease in pressure is known as relative lowering of vapour pressure. Relative lowering of the vapour pressure is equal to the mole fraction of solute. Formula to calculate mole fraction is:
Mole fraction$= \dfrac{{{\text{moles of substance}}}}{{{\text{total moles}}}}$
In this case substance is solute. So, the formula is:
Mole fraction$= \dfrac{{{\text{moles of solute}}}}{{{\text{total moles}}}}$
Moles can be calculated by the formula,
Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$
Given mass of cane sugar is $34.2g$ and molecular mass of cane sugar is $342g$. So, number of moles of cane sugar is,
Number of moles$= \dfrac{{34.2}}{{342}} = 0.1$
Given mass of water is $180g$ and molecular mass of water is $18g$. So, number of moles of water is,
Number of moles$= \dfrac{{180}}{{18}} = 10$
Now, we have moles of solute and total moles (can be calculated by adding the number of moles of cane sugar and water).

So, substituting these values in the formula of calculation of mole fraction,
Mole fraction$= \dfrac{{0.1}}{{0.1 + 10}} = \dfrac{{0.1}}{{10.1}}$
Solving this we get,
Mole fraction$= 0.0099$
As explained above mole fraction of solute is equal to relative lowering of vapour pressure. So, relative lowering of vapour pressure is equal to $0.0099$.
So, the correct answer is option A.

Note:
Colligative properties are those properties which depend on the mole fraction of solute that is the ratio of solute molecules to the total molecules. Relative lowering of vapour pressure is equal to the mole fraction of solute. This means relative lowering of vapour pressure is a colligative property.