Question

Relative lowering of vapor pressure of dilute solution of glucose dissolved in 1 kg of water is 0.002 the morality of the solution is

Answer 1

Hey Dear,

● Answer -

Molarity = 1.12×10^-4 M

● Explanation -

# Given -

∆P/P° = 0.002

Ww = 1 kg

Mw = 18 g/mol

Mg = 180 g/mol

# Solution -

Here, glucose (solute) is dissolved in water (solvent).

No of moles of water can be calculated as -

nw = Ww / Mw

nw = 1000 / 18

nw = 55.56 mol

Relative lowering of vapor pressure is equal to mole fraction of solute.

∆P/P° = Xg = ng / (ng + nw)

0.002 = ng / (ng + 55.56)

0.0002ng + 0.1111 = ng

ng = 0.112 mol

Molarity of the solution is calculated by -

Molarity = ng / Vw

Molarity = 0.112 / 10^-3

Molarity = 1.12×10^-4 M

Hence, molarity of solution is 1.12×10^-4 M.

Hope you understand.

Answer 2

Relative lowering of vapor pressure of dilute solution of glucose dissolved in 1 kg of water is 0.002 the morality of the solution is

:-)