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

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

For 178.2 g water, n_A = 9.9

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

For 18 g glucose, n_B = 0.1

X_B = 0.1/(0.1+9.9) = 0.01

X_A = 0.99

For lowering of vapour pressure,

P = p⁰_AX^A = p⁰_A(1 – X_B)

P = 760(1 – 0.01)

= 760 - 7.6

= 752.4 torr