Question

If 18 g glucose (c6h12o6) is added to 178.2 g water, the vapour pressure of water (in torr) for this aqueous solution is

Answer 1

Weight of glucose = 18g

Molecular weight of glucose = 180g/mol

So, number of mole of glucose = 18/180 = 0.1

Similarly, weight of water = 178.2 g

Molecular weight of water = 18g/mol

So, number of mole of water = 178.2/18 = 9.9

Now, mole fraction of glucose , X  = 0.1/(0.1+9.9) = 0.01

We know, lowering of vapour pressure is directly proportional to mole fraction of solute.

so, ΔP/P₀ = X

⇒ ΔP = P₀X

⇒ P₀ - P = P₀X

⇒ P = P₀(1 - X)

Given, P₀ = 760 torr [ atmospheric pressure]

so, P = 760 (1 - 0.01) = 752.4 torr

Hence, the pressure of water forth is aqueous solution is 752.4 torr