Question

Lowering of vapour pressure of liquid is 1.1 mm
in 0.1m dilute solution. The vapour pressure of
pure liquid is 10x mm. The value of x is
(Molecular mass of liquid is 18 g mol-!) :-
(1) 61.16
(2) 55.5
(3) 93.3
(4) 4.51​

Answer 1

Answer:

The molar mass of water is 18 g/mol

The number of moles of water is 18180=10 mol

The mole fraction of urea is

 X=0.1+100.1=0.0099

The relative lowering in the vapour pressure of solution is equal to the mole fraction of solute

 popo−p=X

 2424−p=0.0099

 24−p=0.2376

 p=23.76 mmHg

Explanation:

the explanation is also in the answer

hope this help u