Question

Dissolving 5 gram of glucose in 75 gram of water will decrease the vapour pressure fron 0,3 atm to....

Answer 1

Mass of glucose: 5 gram
mass of water : 75 gram
Mr of glucose = 180 g/mol
Mr of water = 18 g/mol
P⁰ = 0,3 atm

n glucose = gr/Mr = 5/180 = 1/36 mol
n water = gr/Mr = 75/18 = 150/36 mol
n total = 1/36 + 150/36 = 151/36

Xp = n water/ n total = (150/36) / (151/36) = 150/151

P = P⁰.Xp
= (0,3)(150/151)
= 0,298 atm

Thus, the vapor pressure will decrease from 0,3 atm to 0,298 atm