Respuesta :
Answer: The mass of acetone that must be added is 4.101 grams
Explanation:
The relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.
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 = 1.556 kPa
i = Van't Hoff factor = 1 (for non electrolytes)
[tex]\chi_{solute}[/tex] = mole fraction of solute = ?
[tex]p^o[/tex] = vapor pressure of pure water = 22.022 kPa
Putting values in above equation, we get:
[tex]\frac{1.556}{22.022}=1\times \chi_{acetone}\\\\\chi_{acetone}=0.0707[/tex]
As, the mole fraction of acetone is 0.0707. This means that 0.0707 moles are present in the solution
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of acetone = 58 g/mol
Moles of acetone = 0.0707 moles
Putting values in above equation, we get:
[tex]0.0707mol=\frac{\text{Mass of acetone}}{58g/mol}\\\\\text{Mass of acetone}=(0.0707mol\times 58g/mol)=4.101g[/tex]
Hence, the mass of acetone that must be added is 4.101 grams
