Question

calculate the molality of 1M solution of NaNO3. MOlar mass of NaNO3 id 85. density of the solution is 0.981 g/cm 3

Answer 1

Given,
Density of solution = 0.981 g/cm3
Therefore, mass of 1 L solution = volume x density = 1000 x 0.981 = 981 g
Molar mass of NaNO3 = 85 g
As we have given 1 M solution of NaNO3, it means 1 mole of NaNO3 is present in 1 L volume.
Mass of 1 mole of NaNO3 = 85 g
Mass of solvent = 981 - 85 = 896 g

Molality of NaNO3 = number of moles of solute / mass of solvent in kg
= 1 / 0.896
= 1.116 m

Answer 2

Answer:

1.116 m

Explanation:

Given,

Density of solution = 0.981 g/cm3

Therefore, mass of 1 L solution = volume x density = 1000 x 0.981 = 981 g

Molar mass of NaNO3 = 85 g

As we have given 1 M solution of NaNO3, it means 1 mole of NaNO3 is present in 1 L volume.

Mass of 1 mole of NaNO3 = 85 g

Mass of solvent = 981 - 85 = 896 g

Molality of NaNO3 = number of moles of solute / mass of solvent in kg

= 1 / 0.896

= 1.116 m