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2 years ago

Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

1 answer:

5 0

Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ
1 mol of water weighs = 18.015g
1.55 kg = 1550/18.015 mol = 86.03 mol

Heat required to vaporize :
= 86.03 mol x 40.7 kJ

= 3501.421 kJ

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Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

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Answer: The freezing point of solution is -0.454°C

Explanation:

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,

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

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

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

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

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