Question

5 mol nacl is dissolved in 500g h2o then determine freezing point and boiling point of solution

Answer 1

Answer:

the answer is 270.60k

Explanation:

how can we say means see in the photo

Answer 2

The freezing point of solution is -37.2°C and the boiling point of solution is 110.4°C

Explanation:

• Calculating the freezing point:

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:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Van't hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$n_{solute}$ = Moles of solute (NaCl) = 5 moles

$W_{solvent}$ = Mass of solvent (water) = 500 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{500}\\\\\text{Freezing point of solution}=-37.2^oC$

• Calculating the boiling point:

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:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

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

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure water = 100°C

i = Van't hoff factor = 2

$K_f$ = molal boiling point elevation constant = 0.52°C/m

$n_{solute}$ = Moles of solute (NaCl) = 5 moles

$W_{solvent}$ = Mass of solvent (water) = 500 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.52^oC/m\times \frac{5\times 1000}{500}\\\\\text{Boiling point of solution}=110.4^oC$

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