Answer: Option B
:
B
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.