Question

Calculate the vapour pressure of 10%w/w aqueous glucose sol of 30°c vp of pure water is 31.8

Answer 1

Answer:

Refer attachment

Explanation:

Answer 2

Answer:

Vapour pressure of the glucose solution is 6.36 atm.

Explanation:

10% w/w means 10g of solute in 90g of solvent.

So 10g of glucose in 90g of water.

Number of moles of glucose

Molecular formula of glucose - C₆H₁₂O₆

Molecular weight - (6×12) + (12×1) + (18×6) = 180g

Number of moles $n_a$ - $\frac{10}{180} = \frac{1}{8} = 0.125g$

Number of moles of water

Molecular formula of water - H₂O

Molecular weight of water - 18g

number of moles of water -$\frac{90}{180} = \frac{1}{2} = 0.5g$

So the molar fraction of glucose

$x_A = \frac{n_a}{n_a + n_b} = \frac{0.125}{0.125+0.5} = \frac{0.125}{0.625} = 0.2$

Vapour pressure of a solution = Pressure of pure solvent × mole fraction of solute

V.P = 31.8 ×0.2

V.P = 6.36 atm