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}=(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* \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* 120.9g/mol=1217.463g=1.217kg`

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