Question

1 molar solution of sodium nitrate has density of 1.30 g/cm^-3 .the molality of this solution is​

Answer 1

Given:-

→ Molarity of the solution = 1 M

→ Density of the solution = 1.30 g/cm³

To find:-

→ Molality of the solution.

Solution:-

We are given with the molarity and density of the solution,and we require to calculate the molality.

• Molar mass of Sodium (Na) = 23g/mol

• Molar mass of Nitrogen (N) = 14g/mol

• Molar mass of Oxygen (O) = 16g/mol

Hence, molar mass of NaNO₃ :-

= 23 + 14 + 16×3

= 37 + 48

= 85g/mol

Mass of Sodium Nitrate in 1 L solution :-

= 1 × 85

= 85g

Mass of 1L solution :-

= 1000 × density

= 1000 × 1.30

= 1300g

Mass of solvent in solution :-

= (1300 - 85) g

= 1215g

= 1.215 kg

Molality of a solution :-

= Moles of solute/Mass of solvent in kg

= 1/1.215

= 0.82m

Thus, molality of the solution is 0.82m .