You are provided with 6.33 L of a 4.50 M solution. - id63345011173562, patriciamelton57

Question

Chemistry: You are provided with 6.33 L of a 4.50 M solution. How many moles of solute were required to create this solution

patriciamelton57 · Accepted Answer

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= p_o= 187.54 mmHg

Vapor pressure of the solution at 65 °C= p_s

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

p_s=173.83 mmHg

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

You are provided with 6.33 L of a 4.50 M solution. How many moles of solute were required to create this solution