Question

Ethylene glycol (C2H6O2) is a non-volatile and non-electrolyte compound with a density of 1.11 g.cm–3 at 25.0°C. Calculate the volume ethylene glycol that must be added to 675 cm3 of water to reduce the vapor pressure by 3.82% at 25.0°C. The density and vapor pressure of pure water at 25.0°C are 1.02 g.cm–3 and 23.8 torr respectively

Answer 1

Step 1 : To find the reduced vapor pressure of water.

Vapor pressure of water at 25.0 ⁰C = 23.8 torr.

We have to find the volume of ethylene glycol that must be added to 675 cm³ of water, so as to reduce the vapor pressure by 3.82 % .

New vapor pressure will be equal = 100 - 3.82 = 96.18 % of initial vapor pressure.

96.18 % of 23.8 torr

$= 23.8 torr * \frac{96.18}{100} = 22.891 torr.$

Hence the reduced vapor pressure of water would be 22.89 torr.

Step 2 : To find the mole fraction of water when its vapor pressure is reduced to 22.89 torr.

Let P₁ be the vapor pressure of water in the solution and

P₁⁰ be the vapor pressure of pure water.

Then the vapor pressure of water in the solution is given by the expression

$P_1 = x_1 * P_1 ^0 .$

where x₁ is the mole fraction of water.

substitute P₁ = 22.89 torr and P₁⁰ = 23.8 torr, in the formula we have

$22.89\space torr = x_1 * 23.8\space torr.$

divide both the side by 23.8 torr, we have

$x_1 = \frac{22.89\space torr }{23.8\space torr} = 0.9618 .$

Step 3 : To find the moles of water in the solution.

volume of water = 675 cm³ .

density of water = 1.02 g/cm³ .

plug volume and density of water in the density formula and find the mass of the water.

$density = \frac{mass}{volume } ;$

arranging this formula for mass we have

$mass = density * volume = 1.02\space g/cm^3 * 675\space cm^3 = 688.5\space grams.$

molar mass of water = 18.0148 g/mol.

$$moles\space of\space water = 688.5\space g\space of\space H_2O * \frac{1\space mol\space of \space H_2O}{18.0148\space g\space of\space H_2O}$$

$= 38.219\space mol\space of \space H_2O$

Step 4 : To find the moles of moles of ethylene glycol.

$$mole\space fraction\space of\space water = \frac{mol\space of \space water }{moles\space of\space solution }$$ $$total \space moles \space in \space solution = \frac{mol \space of \space water }{mole \space fraction \space of \space water }$$

$total \space moles \space in \space solution = \frac{38.219\space mol }{0.9618 } = 39.737\space moles$

moles of ethylene glycol = moles of solution - moles of solvent

= 39.737 mol - 38.219 mol

= 1.5180 moles of ethylene glycol.

Step 5 : To convert 1.518 mol of ethylene glycol to grams.

molar mass of ethylene glycol = 62.0674 g/mole

grams of ethylene glycol = moles * molar mass

= 1.5180 mol * 62.0674 g/mol

= 94.2121 grams.

Step 6 : To convert 94.2121 grams of ethylene glycol to 'mL'

volume = mass/ density = 94.2121/ 1.11 g/mL

= 84.89 mL

Which in 3 significant figure is 84.9 mL

Hence we must add 84.9 mL of ethylene glycol to 675 mL of water, so as to reduce the vapor pressure by 3.82 % of initial vapor pressure of water.