Question

A particular refrigerant cools by evaporating liquefied CCl 2F 2. How many kg of the liquid must be evaporated to freeze a tray of water to ice (at zero degrees C)? The tray contains 525 grams water. Molar heat of fusion of ice = 6.01 kJ/mol. Molar heat of vaporization of CCl 2F 2 = 17.4 kJ/mole

Answer 1

Answer : The mass of $CCl_2F_2$ evaporated must be, 1.217 kg

Explanation :

First we have to calculate the moles of water.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}$

Molar mass of water = 18 g/mol

$\text{Moles of water}=\frac{525g}{18g/mol}=29.17mol$

Now we have to calculate the heat released.

Heat released = Moles of water × Molar heat of fusion of ice

Heat released = 29.17 mol × 6.01 kJ/mol

Heat released = 175.3 kJ

Now we have to calculate the moles of $CCl_2F_2$

Heat = Moles of $CCl_2F_2$ × Molar heat of vaporization of $CCl_2F_2$

175.3 kJ = Moles of $CCl_2F_2$ × 17.4 kJ/mol

Moles of $CCl_2F_2$ = 10.07 mol

Now we have to calculate the mass of $CCl_2F_2$

$\text{Mass of }CCl_2F_2=\text{Moles of }CCl_2F_2\times \text{Molar mass of }CCl_2F_2$

Molar mass of $CCl_2F_2$ = 120.9 g/mol

$\text{Mass of }CCl_2F_2=10.07mol\times 120.9g/mol=1217.463g=1.217kg$

Thus, the mass of $CCl_2F_2$ evaporated must be, 1.217 kg