Question

18 GM's of glucose is added to 178.2 g water the vapour pressure of water(in torr) for this aqueous solution is???​

Answer 1

Answer:

18g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is:

76.0

752.4

759.0

7.6

Solution

Molecular mass of water = 2×1 + 1×16 = 18 g

For 178.2 g water, nA = 9.9

Molecular mass of glucose = 6×12 + 12×1 + 6×16 = 180 g

For 18 g glucose, nB = 0.1

XB = 0.1/(0.1+9.9) = 0.01

XA = 0.99

For lowering of vapour pressure,

P = p0AXA = p0A(1 – XB)

P = 760(1 – 0.01)

= 760 - 7.6

= 752.4 torr