I really cant read it sorry i tried
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
Explanation:
It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.
Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.
So, vapor pressure of mixture = 1% vapor pressure of pure water
Therefore,
=
= 27.785 Pa
Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.