Ncert Exercises & Solutions

The density of 3 molal solution of NaOH is 1.110 g mL^-1. Calculate the molarity of the solution.

3 molal solution of NaOH means 3 moles of NaOH are dissolved in 1 kg solvent. So, the mass of solution = 1000 g solvent + 120 g NaOH = 1120 g solution

(Molar mass of NaOH

= 23 + 16 + 1 = 40 g

and 3 moles of NaOH

= 3 \times 40 = 120 g

)

Volume of solution = \frac{Mass of solution}{Density of solution}

V = \frac{1120 g}{1.110 g {mL}^{- 1}} = 1009 mL

Molarity = \frac{Moles of solute \times 1000}{Volume of solution (mL)}

= \frac{3 \times 1000}{1009} = 2.973 M \approx 3 M

Hint: Use the formula of molarity = \(\frac{number \ of \ mole \ solute}{volume \ of\ solution (L)}\)

Step 1:

3 molal solution of NaOH means 3 moles of NaOH are dissolved in 1 kg solvent.

So, the mass of solution = 1000 g solvent + 120 g NaOH = 1120 g solution

(Molar mass of NaOH

= 23 + 16 + 1 = 40 g

and 3 moles of NaOH

= 3 \times 40 = 120 g

)

Step 2:

Volume of solution = \frac{Mass of solution}{Density of solution}

V = \frac{1120 g}{1.110 g {mL}^{- 1}} = 1009 mL

Molarity = \frac{Moles of solute \times 1000}{Volume of solution (mL)}

= \frac{3 \times 1000}{1009} = 2.973 M \approx 3 M

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions