Respuesta :

Answer: The freezing point of solution is -10.9°C and the boiling point of solution is 103.°C

Explanation:

To calculate the mass of water, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of water = 1 g/mL

Volume of water = 129 mL

Putting values in above equation, we get:

[tex]1g/mL=\frac{\text{Mass of water}}{129mL}\\\\\text{Mass of water}=(1g/mL\times 129mL)=129g[/tex]

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

  • The equation used to calculate depression in freezing point follows:

[tex]\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}[/tex]

To calculate the depression in freezing point, we use the equation:

[tex]\Delta T_f=iK_fm[/tex]

Or,

[tex]\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2 (For NaCl)

[tex]K_f[/tex] = molal freezing point elevation constant = 1.86°C/m

[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 22.1 g

[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 58.5  g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 129 g

Putting values in above equation, we get:

[tex]0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{22.1\times 1000}{58.5g/mol\times 129}\\\\\text{Freezing point of solution}=-10.9^oC[/tex]

Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.

  • The equation used to calculate elevation in boiling point follows:

[tex]\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}[/tex]

To calculate the elevation in boiling point, we use the equation:

[tex]\Delta T_b=iK_bm[/tex]

Or,

[tex]\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For NaCl)

[tex]K_b[/tex] = molal boiling point elevation constant = 0.512°C/m

[tex]m_{solute}[/tex] = Given mass of solute (NaCl) = 22.1 g

[tex]M_{solute}[/tex] = Molar mass of solute (NaCl) = 58.5  g/mol

[tex]W_{solvent}[/tex] = Mass of solvent (water) = 129 g

Putting values in above equation, we get:

[tex]\text{Boiling point of solution}-100=2\times 0.512^oC/m\times \frac{22.1\times 1000}{58.5g/mol\times 129}\\\\\text{Boiling point of solution}=103.^oC[/tex]

Hence, the freezing point of solution is -10.9°C and the boiling point of solution is 103.°C