Question

Calculate the vapor pressure lowering caused by addition of 50g of sucrose (molecular mass= $342$ to $500g$ of water if the vapor pressure of pure water at ${25^ \circ }C$ is $23.8mmHg$ .

Answer 1

Hint: Vapor pressure is defined as the pressure that is exerted by the vapors over the liquid under equilibrium conditions at a given temperature. It is given by the formula: ${P_{solution}} = {P_{solvent}}{X_{solvent}}$ .

Complete step by step answer:
Vapor pressure lowering or relative lowering of vapor pressure is defined as the vapor pressure of solvent that is present in solution is less than the vapor pressure of the pure solvent.
It is given by the formula: Relative lowering of vapor pressure $ = \dfrac{{\Delta p}}{{p_1^0}}$………….(1
)
Where,
$\Delta p = $ Vapor pressure of solution.
$p_1^0 = $ Vapor pressure of a pure solvent.
Relative lowering of vapor pressure $ = \dfrac{{p - {p_s}}}{p}$
Where, $p = $ vapor pressure of solvent.
${p_s} = $ Vapor pressure of pure solvent.
Mathematically according to Raoult’s law relative lowering of vapor pressure is given as follows:
$\dfrac{{p - {p_s}}}{p} = \dfrac{n}{{n + N}}$ …………….….(2)
Where $N = $Number of moles of solvent
$n = $Number of moles of solute.
First we will calculate the number of moles of sucrose and water respectively.
Number of moles is defined as the ratio of mass to the molar mass. It is given by the formula $n = \dfrac{m}{M}$
Where,
$n = $ Number of moles
$m = $ Given mass
$M = $ molecular mass
Given data: molecular mass of sucrose $\left( {{M_1}} \right) = 342$
Mass of sucrose $\left( {{m_1}} \right) = 50g$
Molecular mass of water $\left( {{M_2}} \right) = 18$
Mass of water $\left( {{m_2}} \right) = 500g$
Number of moles of sucrose ${n_{}} = \dfrac{{{m_1}}}{{{M_1}}}$
Substituting the values we get,
${n_{}} = \dfrac{{50}}{{342}}$
${n_{}} = 0.146$ moles
Number of moles of water $N = \dfrac{{{m_2}}}{{{M_2}}}$
Substituting the values we get,
$N = \dfrac{{500}}{{18}}$
$N = 27.78$ moles

Now using the formula of relative lowering of vapor pressure we will find out the vapor pressure lowering caused due to addition of sucrose.
Given data:
Vapor pressure of pure water ${p_0} = 23.8mmHg$
${n_{}} = 0.146$
$N = 27.78$
Combining equation (1) and equation (2) we get,
$\dfrac{{\Delta p}}{{p_1^0}} = \dfrac{{p - {p_s}}}{p}$
Then,
$\dfrac{{\Delta p}}{{p_0^{}}} = \dfrac{n}{{n + N}}$
Therefore,
$\Delta p = \dfrac{n}{{n + N}}{p_0}$
Substituting the values we get,
$\Delta p = \dfrac{{0.146}}{{0.146 + 27.78}} \times 23.80$
$\Delta p = \dfrac{{0.146}}{{27.926}} \times 23.80$
$\Delta p = 0.0052 \times 23.80$
$\Delta p = 0.124mmHg$

Therefore, the vapor pressure that is caused by addition of sucrose is $0.124mmHg$ .

Note: Relative lowering of vapor pressure is a colligative property. Vapor pressure of solution is dependent on the number of moles of solvent present in the solution.