Question

If 0.340 mol of a nonvolatile nonelectrolyte are dissolved in 3.10 mol of water, what is the vapor pressure PH2O of the resulting solution? The vapor pressure of pure water is 23.8 torr at 25 ∘C . Express your answer with the appropriate units.

Answer 1

Answer:

21.4 torr

Explanation:

The vapour pressure of a solution is lower than that of the pure solvent. And according to Raoult's law this depends of the amount of solute in the solvent. It is calculated according to the following equation

$P_{A} =X_{A} P^{o} _{A}$

where $P_{A} is the pressure of the solution\\ X_{A} is the mole fraction of the solvent\\ P^{o}_{A}is the mole fraction of pure solvent$

$P_{A}=\frac{3.10 mol}{3.10mol+0.340mol}*23.8torr=21.4torr$

Answer 2

The vapour pressure is exerted by the vapours of the substances in a closed environment at a provided temperature. The vapour pressure of the resulting solution is 21.4 torr.

What is vapour pressure?

Raoult's law states that the vapour pressure of a solution is less than pure solvent and relies on the quantity of solute dissolved in the solvent.

The vapour pressure can be calculated as:

$\rm P_{A} = X_{A} P^{\circ}_{A}$

Where,

$\rm P_{A}$ = Pressure of solution

$\rm X_{A}$ = Mole fraction of the solvent

$\rm P^{\circ}_{A}$ = Mole fraction of pure solvent

Given,

• The vapour pressure of pure water = 23.8 torr
• Moles of nonvolatile nonelectrolyte = 0.340 moles
• Moles of the water =  3.10 mol

Substituting values in the above equation we get:

$\begin{aligned} \rm P_{A} &=\dfrac{3.10}{3.10+0.340} \times 23.8 \\\\&= 21.4 \;\rm torr \end{aligned}$

Therefore, 21.4 torr is the vapour pressure of the solution.

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