Question

Vapour pressure of water at 293 k is 17.535 mmhg vapour pressure of water at 293 K when 25 gram glucose dissolved in 450 gram of water

Answer 1

your ans should be............

Vapour pressure of water, p1° = 17.535 mm of Hg

Mass of glucose, w2 = 25 g

Mass of water, w1 = 450 g

We know that,

Molar mass of glucose (C6H12O6),

M2 = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1

Molar mass of water, M1 = 18 g mol - 1

Then, number of moles of glucose, n1 = 25/180 = 0.139 mol

And, number of moles of water, n2 =450/18 = 25 mol

Now, we know that,

(p1° - p°) / p1° =  n1 / n2 + n1

⇒ 17.535 - p°  / 17.535 =   0.139 / (0.139+25)

⇒ 17.535 - p1 = 0.097

⇒ p1 = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.