Question

Shikhar was to perform a neutralization reaction in laboratory. In order to neutralize 500 ml of $2MNaOH$ solution he was supposed to add certain volume of $1M{H}_{3}P{O}_3$ solution, but by mistake, he added the same volume of water instead. What is the molality of the resulting solution – Assume density of the original solution = $1.08g/cc$ Density of water = 1 g/cc.

• A. 0.67
• B. 1
• C. 1.2
• D. 2

Answer 1

The correct option is B

1

Wt. of original solution = Density x volume = $1.08g/cc\times 500ml=540g$

Wt. of NaOH + = No. of moles x molecular Wt. = $\frac{500\times 2}{1000}\times 40g=40g$

Wt. of solvent i.e water in original solution = 540-40 = 500 g

if $V_2$ be the volume of $H_{3}P{O}_3$ required for neutralizing the solution then-

$M_1{V}_1=M_2{V}_2$

$\Rightarrow V_2=\frac{M_1{V}_1}{M_2}=\frac{2\times 500}{1\times 2}ml=500ml$

Note-Basicity of $H_{3}P{O}_3=2$

Wt. of water added = Density of water $\times$ Volume = $\frac{1g}{cc}\times 500ml=500g$

Total Wt. of solvent i.e water = 500 + 500 = 1000g

Molality of solution = $\frac{Molesofsolute}{Wt.ofsolventinkg}=\frac{40/40}{1000/1000}=1mole/kg$

Hence Optin-(b) is the correct Answer.