Question

An aqueous solution containing glucose has a vapor pressure of 17.1 torr at 25 degrees C. What would be the vapor pressure of this solution at 45 degrees C? The vapor pressure of pure water is 23.8 torr at 25 degrees C and 71.9 torr at 45 degrees C. If the glucose in the solution were substituted with an equivalent amount (moles) of NaCl what would be the vapor pressure at 45 degrees C?

Answer 1

By Raoult's law

Relative lowering of vapor pressure is

(P^o - P)/P^o = X₂

At , 25 degree

Vapor pressure at pure water (P^o ) = 23.8 torr

Vapor pressure of glucose solution ( P ) = 17.1 torr.

X₂ mole fraction of solute.

So, [ (23.8 - 17.1)/23.8 ] = X₂

Or, X₂ = 0.2815

Now, at 45 degree

P^o = 71.9 torr

So, (P^o - P)/P^o = 0.2815

Or, [ ( 71.9 - P)/71.9] = 0.2815

Or, 71.9 - P = 20.24

Or, P = 51.66 torr.

Vapor pressure of glucose solution at 45 degree is 51.67 torr.

Part 3

NaCl is an electrolyte , it dissociates to Na⁺and Cl^-.

Then , vant hoff factor (i) = 2

Then,

[(P^O- P)/P⁰ ] = i * X₂

Or, [(71.9 - P)/71.9] = 2* 0.2815

Or, (71.9 - P) = 2*0.2815*71.9

Or (71.9- P) = 40.48

Or, P = (71.9 - 40.48) = 31.42 torr.

So, vapor pressure of same amount of NaCl solution at 45 degree is 31.42 torr.