Respuesta :
Answer: The Van't Hoff factor for KCl is 1.74
Explanation:
We are given:
Molality of solution = 1 m
This means that 1 mole of a solute is present in 1 kg of solvent (water) or 1000 grams of water
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of water = 1000 g
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol[/tex]
Mole fraction of a substance is given by:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex]
Moles of solute = 1 moles
Total moles = [1 + 55.56] = 56.56 moles
Putting values in above equation, we get:
[tex]\chi_{(solute)}=\frac{1}{56.56}=0.0177[/tex]
The equation used to calculate relative lowering of vapor pressure follows:
[tex]\frac{p^o-p_s}{p^o}=i\times \chi_{solute}[/tex]
where,
[tex]\frac{p^o-p_s}{p^o}[/tex] = relative lowering in vapor pressure = 0.734 mmHg
i = Van't Hoff factor = ?
[tex]\chi_{solute}[/tex] = mole fraction of solute = 0.0177
[tex]p^o[/tex] = vapor pressure of pure water = 23.76 torr
Putting values in above equation, we get:
[tex]\frac{0.734}{23.76}=i\times 0.0177\\\\i=1.74[/tex]
Hence, the Van't Hoff factor for KCl is 1.74
The Van't Hoff factor for the KCl solution is 1.72.
We are told in the question that the solution is 1 m, this means that it contains 1 mole of KCl in 1 Kg of water.
Number of moles of water = 1000 g/18 g/mol = 55.56 moles
Mole fraction of the solute = 1/1 + 55.56 = 0.018
Using the formula for vapor pressure lowering;
p° - ps/p° = i Χ
Where;
p° = vapor pressure of pure solvent
ps = vapor pressure of solution
i = Van't Hoff factor
Χ = mole fraction
Since p° - ps = 0.734 mmHg
0.734 mmHg/23.76 mmHg = 0.018 i
i = 1.72
The Van't Hoff factor for the KCl solution is 1.72.
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