Question 53 E - Chemistry (OpenStax)

Question

A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance. The Vapour Pressure of the Solution is 23.35 Tor.

Step-by-Step Solution

Answer

 The Molar Mass of the Non-Electrolyte is 144 g/mole. 

Step 1: Vapour Pressure

Vapour pressure may be defined as the pressure when the liquid and vapour are at equilibrium. The equilibrium force per unit area is understood to function an indicator of the evaporation rate of a liquid.

Step 2: Necessary Steps

Mass of Non-Electrolyte= 12.0g

Molar Mass of Non-Electrolyte= x g/mole

Mass of Water= 80g= 0.80g

Molar Mass of Water= 18 g/mole

 

Firstly,

 \(\begin{aligned}{l}{\bf{Molality }} &= \dfrac{{{\bf{Number}}\,{\bf{of}}\,{\bf{moles}}}}{{{\bf{Mass}}\,{\bf{of}}\,{\bf{Solvent(kg)}}}}\\{\bf{Molality }}&= \dfrac{{{\bf{12/x}}}}{{{\bf{0}}{\bf{.080}}}}\end{aligned}\)

 

\({\bf{\Delta Tfreezing = i \times Kf \times m}}\)

Step 3: Explanation

Vapour Pressure of Solution= 23.35 Tor

\({{\bf{k}}_{\bf{f}}}\) value of Water= 1.86

\(\begin{aligned}{l}{\bf{\Delta Tf}} & = {\rm {i \times Kf \times m}}\\{\bf{0 - Tf}} &= {\rm{ 1 \times 1}}{\bf{.86 \times }}\frac{{{\bf{12/x}}}}{{{\bf{0}}{\bf{.080}}}}\\{\bf{0 - ( - 1}}{\bf{.94)} & = {\rm{ 1}}{\bf{.86 \times }}\frac{{{\bf{12/x}}}}{{{\bf{0}}{\bf{.080}}}}\\{\bf{1}}{\bf{.94/1}}{\bf{.86 }} &= \dfrac{{{\bf{12/x}}}}{{{\bf{0}}{\bf{.080}}}}\\{\bf{1}}{\bf{.043 \times 0}}{\bf{.080}} & = {\rm {12/x}}\\{\bf{x }} &= \dfrac{{{\bf{12 \times 1}}{\bf{.043}}}}{{{\bf{0}}{\bf{.080}}}}\\{\bf{x}}& = {\rm{144g/mole}}\end{aligned}\)

 

The Molar Mass of Non-Electrolyte=144 g/mole