The vapour pressure of pure water at 110 c is 1070 torr.A soln of ethyleneglycol in water has vapour pressure of 1 atm. Assuming that raoult's law is obeyed, what is the mole fraction of ethylene glycol in water?
0.290
chemistry
To find the mole fraction of ethylene glycol in water, we can use Raoult's Law, which states that the vapor pressure of a solvent above a solution is proportional to the mole fraction of the solvent in the solution.
According to Raoult's Law, the vapor pressure of water in the solution is given by:
P_water = X_water * P_water^pure
Where:
P_water is the vapor pressure of water in the solution
X_water is the mole fraction of water in the solution
P_water^pure is the vapor pressure of pure water
In this case, we are given:
P_water = 1 atm
P_water^pure = 1070 torr (which is equivalent to 1.4 atm)
We need to convert the units of pressure to be consistent, so we convert 1070 torr to atm:
1.4 atm = 1070 torr * (1 atm/760 torr)
Now we can rearrange the equation to solve for X_water:
X_water = P_water / P_water^pure
Substituting the values we have:
X_water = 1 atm / 1.4 atm
Therefore, the mole fraction of water (X_water) is approximately 0.714.
To find the mole fraction of ethylene glycol, we can use the fact that the sum of mole fractions in a solution is always equal to 1:
X_ethylene_glycol = 1 - X_water
Substituting the value of X_water we found:
X_ethylene_glycol = 1 - 0.714
Therefore, the mole fraction of ethylene glycol (X_ethylene_glycol) is approximately 0.286.