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ira [324]
3 years ago
13

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

mole(s) of CsF(s) in 1.00 kg of water? Assume that Raoult's law applies.
Chemistry
1 answer:
luda_lava [24]3 years ago
5 0

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of H_2O = 1.00 kg = 1000 g

Moles of CsF = 3.68 mole

Molar mass of H_2O = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of H_2O.

\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole

Now we have to calculate the mole fraction of H_2O

\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938

Now we have to partial pressure of solution.

According to the Raoult's law,

P_{Solution}=X_{H_2O}\times P^o_{H_2O}

where,

P_{Solution} = vapor pressure of solution

P^o_{H_2O} = vapor pressure of water = 0.692 atm

X_{H_2O} = mole fraction of water = 0.938

P_{Solution}=X_{H_2O}\times P^o_{H_2O}

P_{Solution}=0.938\times 0.692atm

P_{Solution}=0.649atm

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

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