Question

An aqueous solution is 12.0 % by mass silver nitrate, AgNO₃, and has a density of 1.11 g/mL.

The mole fraction of silver nitrate in the solution is ________

.

.

Answer 1

let mass of solution be 100 g

Then,

mass of AgNO3 = 12% of 100 g = 12 g

mass of water = 100 g - 12 g = 88 g

Molar mass of AgNO3,

MM = 1*MM(Ag) + 1*MM(N) + 3*MM(O)

= 1*107.9 + 1*14.01 + 3*16.0

= 169.91 g/mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

n(AgNO3) = mass/molar mass

= 12.0/169.91

= 0.070626

n(H2O) = mass/molar mass

= 88.0/18.016

= 4.884547

n(AgNO3),n1 = 0.070626 mol

n(H2O),n2 = 4.884547 mol

Total number of mol = n1+n2

= 0.070626 + 4.884547

= 4.955173 mol

Mole fraction of each components are

X(AgNO3) = n1/total mol

= 0.070626/4.955173

= 0.0143

Answer: 0.0143