Question

An aqueous solution of $NaOH$ having density $1.1$ $kg/{dm}^3$ contains $0.02$ mole fraction of $NaOH$. The molality and molarity of $NaOH$ solution respectively, are :

• A. $0.986,1.134$
• B. $1.134,1.193$
• C. $1.134,1.02$
• D. $1.034,1.134$

Answer 1

Answer: (B)

$1.134,1.193$

The molar masses of NaOH and water are 40 g/mol and 18 g/mol respectively.

Assume, total number of moles of NaOH and water is 1.

The mole fraction of water $=1-0.02=0.98$

$0.02$ moles of NaOH corresponds to $0.02\times 40=0.8$ g.

The number of moles of water is 0.98.

The mass of water $=0.98\times 18=17.64$ g.

Total mass of the solution $=17.64+0.8=18.44$g.

The density of the solution is 1.1 kg/dm${}^3$ or 1.1 g/mL.

The volume of the solution $=\frac{18.44}{1.1}=16.76$ mL.

The molarity of the solution is the number of moles of NaOH present in 1 L of solution.

$\therefore$ The molarity of the solution $=\frac{0.02}{\frac{16.76}{1000}}=1.193M$

The molality of the solution is the number of moles of NaOH present in 1 kg of water.

$\therefore$ The molality of the solution $=\frac{0.02}{\frac{17.64}{1000}}=1.134M$