Chapter 14.4, Problem 52SSC

Interpretation Introduction

Interpretation:

The boiling point elevation and freezing point depression of a solution containing 50 g of glucose in 500 g of water needs to be determined.

Concept introduction:

Boiling point elevation is calculated as follows:

$\Delta {\mathrm{T}}_b=K_b\times m$

Here,

$\Delta T_b=T_{b}ofthesolution-T_{b}ofthesolvent(Water)$

Depression of freezing point is calculated as follows:

$\Delta T_f=K_f\times m$

Molality of solution is calculated as follows:

$\mathrm{Molality}ofsolution=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{mass of solvent in kg}}$

Also,

$\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Answer to Problem 52SSC

The boiling point elevation and freezing point depression of a solution containing 50 g of glucose in 500 g of water are 0.285°C and 1.03°Crespectively.

Boiling point elevation $\Delta T_b=K_b\times m$

Data given:

Mass of glucose=50g

Mass of water=500g

$\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}$

Again, $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Number of moles of glucose $=\frac{Massofglucose}{Molecularweightofglucose}$

$=\frac{50}{180.15(Molecularweightofglucose=180.15)}$

$\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}$

$=\frac{50\times 1000(kg)}{180.15\times 500}$

=0.556m

$\Delta T_f=1.86\times 0.556$

K_b of water = 0.512 $°C/m$

$\Delta T_b=K_b\times m$

$\Delta T_b=0.512\times 0.556$

=0.285°C

Depression of freezing point $\Delta T_f=K_f\times m$

molality of the solution = 0.556 m

K_f of water = 1.86 $°C/m$

$\Delta T_f=K_f\times m$

$\Delta T_f=1.86\times 0.556$

=1.03°C

The boiling point elevation and freezing point depression of a solution containing 50 g of glucose in 500 g of water are 0.285°C and 1.03°C respectively.

Explanation of Solution

Boiling point elevation $\Delta T_b=K_b\times m$

Data given:

Mass of glucose=50g

Mass of water=500g

$\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}$

Again, $\mathrm{Number}ofmolesofsolute=\frac{Massofsolute}{molecularweightofsolute}$

Number of moles of glucose $=\frac{Massofglucose}{Molecularweightofglucose}$

$=\frac{50}{180.15(Molecularweightofglucose=180.15)}$

$\mathrm{Molality}ofsolution=\frac{Numberofmolesofsolute}{Molecularweightofsolute}$

$=\frac{50\times 1000(kg)}{180.15\times 500}$

=0.556m

$\Delta T_f=1.86\times 0.556$

K_b of water = 0.512 $°C/m$

$\Delta T_b=K_b\times m$

$\Delta T_b=0.512\times 0.556$

=0.285°C

Depression of freezing point $\Delta T_f=K_f\times m$

molality of the solution = 0.556 m

K_f of water = 1.86 $°C/m$

$\Delta T_f=K_f\times m$

$\Delta T_f=1.86\times 0.556$

=1.03°C

The boiling point elevation and freezing point depression of a solution containing 50 g of glucose in 500 g of water are 0.285°C and 1.03°C respectively.