Question

A solution contains \(0.115 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(30^{\circ} \mathrm{C}\) is \(25.7\) torr. The vapor pressure of pure water at this temperature is \(31.8\) torr. Calculate the number of moles of sodium chloride in the solution. (Hint: remember that sodium chloride is a strong electrolyte.)

Short answer

The number of moles of sodium chloride in the solution, taking into account its dissociation as a strong electrolyte, is 0.0546 moles.

Step-by-step solution

Use Raoult's Law

Raoult's Law states that the vapor pressure of a solvent (P) in an ideal solution is equal to the vapor pressure of the pure solvent (P°) multiplied by its mole fraction (X) in the solution: P = X * P° In this problem, we have the vapor pressure of the solution (P) and the vapor pressure of pure water (P°). We can use Raoult's Law to find the mole fraction of water in the solution.

Calculate the mole fraction of water in the solution

Rearrange the Raoult's Law formula to find the mole fraction of water in the solution (X): X = P / P° X = 25.7 torr / 31.8 torr X = 0.808 We found that the mole fraction of water in the solution is 0.808.

Calculate the mole fraction of sodium chloride in the solution

To calculate the mole fraction of sodium chloride in the solution, subtract the mole fraction of water from 1: Mole fraction of sodium chloride = 1 - X Mole fraction of sodium chloride = 1 - 0.808 Mole fraction of sodium chloride = 0.192 The mole fraction of sodium chloride in the solution is 0.192.

Find the total moles in the solution

Let's designate the moles of sodium chloride as x. According to the definition of mole fraction, we have: Mole fraction of NaCl = Moles of NaCl / (Moles of NaCl + Moles of H2O) 0.192 = x / (x + 0.115) Rearrange the equation to solve for x: x = 0.192 * (x + 0.115) x = 0.192 * x + 0.192 * 0.115 x - 0.192x = 0.192 * 0.115 0.808x = 0.02208 x = 0.0273 The number of moles of sodium chloride in the solution is 0.0273. However, since sodium chloride is a strong electrolyte, it dissociates into two ions: Na+ and Cl-.

Calculate the total moles of sodium chloride accounting for dissociation

Taking into account the dissociation of sodium chloride, we'll have twice the amount of moles: Moles of NaCl = 2 * 0.0273 = 0.0546 moles So, there are 0.0546 moles of sodium chloride in the solution, accounting for its dissociation.