Question

relative decrease in vapour pressure of an aqueous NaCl is 0.167 number of moles of NaCl present in 180 gram of H2o is

Answer 1

it's really nice question. here you should use two concepts lowering of vapor pressure and Van't Hoff's concept.

i.e., lowering of vapor pressure = mole fraction of solute

= i n/(in + iN)

where i is Van't Hoff's factor, n is number of mole of solute and N is number of mole of solvent .

given, mass of solvent = 180gm

molar mass of solvent = 18g/mol [as solvent is water ]

so, N = 180/18 = 10

Van't Hoff's factor , i = 2 [ NaCl dissociates into two ions ]

now, 0.167 = 2n/(2n + 10)

or, 0.167(2n + 10) = 2n

or, 0.334n + 1.67 = 2n

or, (2 - 0.334)n = 1.67

or, n = 1.67/(1.666) ≈ 1 mol

hence, answer is 1 mol