Question

When a specific amount of acetone (C3H6O) is added to 100.0 g of pure water at 65°C, the vapor pressure of water over the solution is lowered by 1.556 kPa. Given the vapor pressure of water at 65°C is 25.022 kPa, what is the mass of acetone added?

Answer 1

Answer: The mass of acetone that must be added is 4.101 grams

Step-by-step 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:

`(p^o-p_s)/(p^o)=i* \chi_(solute)`

where,

`(p^o-p_s)/(p^o)` = relative lowering in vapor pressure = 1.556 kPa

i = Van't Hoff factor = 1 (for non electrolytes)

`\chi_(solute)` = mole fraction of solute = ?

`p^o` = vapor pressure of pure water = 22.022 kPa

Putting values in above equation, we get:

`(1.556)/(22.022)=1* \chi_(acetone)\\\\\chi_(acetone)=0.0707`

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:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of acetone = 58 g/mol

Moles of acetone = 0.0707 moles

Putting values in above equation, we get:

`0.0707mol=\frac{\text{Mass of acetone}}{58g/mol}\\\\\text{Mass of acetone}=(0.0707mol* 58g/mol)=4.101g`

Hence, the mass of acetone that must be added is 4.101 grams