Question

Consider a perfectly insulated and sealed container. Determine the minimum volume in L of the container such that 0.37 L of water will completely evaporate at 25oC. The heat of vaporization of water is 42.68 kJ/mol, and the density of water is 1.00 g/mL. Report your answer to 2 decimal places. g

Answer 1

Answer: The minimum volume of the container must be 460.54 L.

Explanation:

Density is defined as the ratio of mass and volume of a substance.

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$           ......(1)

Given values:

Volume of water = 0.37 L = 370 mL               (Conversion factor: 1 L = 1000 mL)

Density of water = 1.00 g/mL

Putting values in equation 1, we get:

$\text{Mass of water}=(1.00g/mL\times 370mL)=370g$

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of water = 370 g

Molar mass of water = 18 g/mol

Putting values in equation 2, we get:

$\text{Moles of water}=\frac{370g}{18g/mol}=20.56mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

Applying unitary method:

20.56 moles of water will occupy = $\frac{22.4L}{1mol}\times 20.56mol=460.54L$ of volume.

Hence, the minimum volume of the container must be 460.54 L.