A solution is prepared by adding 1.50 mol glucose, which is not volatile, to 3.50 mol diethyl ether. What is the vapor pressure of this solution at 25 $\circ$ C given that the vapor pressure of pure diethyl ether is 23.8 torr?
A)7.00 torr
B)16.7 torr
C)10.2 torr
D)7.14 torr
E)34.0 torr

Question

Quizplus · Accepted Answer

We can use the following equation to calculate the vapor pressure of the solution:

P solution = X solvent * P° solvent

where X solvent is the mole fraction of the solvent (diethyl ether) and P° solvent is the vapor pressure of pure solvent.

We can first calculate the mole fraction of diethyl ether:

X diethyl ether = moles diethyl ether / total moles 
X diethyl ether = 3.50 / (3.50 + 1.50) 
X diethyl ether = 0.700

Therefore, the mole fraction of glucose is:

X glucose = 1.00 - X diethyl ether 
X glucose = 0.300

Now we can calculate the vapor pressure of the solution:

P solution = X diethyl ether * P° diethyl ether 
P solution = 0.700 * 23.8 torr 
P solution = 16.7 torr

So the answer is B, 16.7 torr.

A Solution Is Prepared by Adding 1 ∘\circ∘ C Given That the Vapor Pressure of Pure Diethyl Ether

A Solution Is Prepared by Adding 1 $\circ$ C Given That the Vapor Pressure of Pure Diethyl Ether