Shikhar was to perform a neutralization reaction in laboratory. In order to ne

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Shikhar was to perform a neutralization reaction in laboratory. In order to neutralize 500 ml of $2 M N a O H$ solution he was supposed to add certain volume of $1 M H_{3} P O_{3}$ 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 = $1.08 g / c c$ Density of water = 1 g/cc.

Options:

A . 0.67

B . 1

C . 1.2

D . 2

Answer: Option B
:
B

Wt. of original solution = Density x volume = $1.08 g / c c \times 500 m l = 540 g$
Wt. of NaOH + = No. of moles x molecular Wt. = $\frac{500 \times 2}{1000} \times 40 g = 40 g$
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-
$M_{1} V_{1} = M_{2} V_{2}$
$\Rightarrow V_{2} = \frac{M_{1} V_{1}}{M_{2}} = \frac{2 \times 500}{1 \times 2} m l = 500 m l$
Note-Basicity of $H_{3} P O_{3} = 2$
Wt. of water added = Density of water $\times$ Volume = $\frac{1 g}{c c} \times 500 m l = 500 g$
Total Wt. of solvent i.e water = 500 + 500 = 1000g
Molality of solution = $\frac{M o l e s o f s o l u t e}{W t . o f s o l v e n t i n k g} = \frac{40 / 40}{1000 / 1000} = 1 m o l e / k g$
Hence Optin-(b) is the correct Answer.

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