Question

The molality (m) of each ion $(N{H}_4^{+}$ and $C{l}^{-})$, respectively, present in the aqueous solution of $2MN{H}_{4}Cl$ assuming $100\%$ dissociation according to given reaction is,
$N{H}_{4}Cl\left(aq\right)\to N{H}_4^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

Given that density of solution $=3.107$ g/ml.

• A. $0.45m,0.67m$
• B. $0.67m,0.45m$
• C. $0.67m,0.67m$
• D. $0.45m,0.45m$

Answer 1

Answer: (C)

$0.67m,0.67m$

Since the dissociation is $100\%$, the molarity of each ion is equal to the molarity of ammonium chloride solution before dissociation which is $2M$.
Thus, $1L$ of the solution contains $2$ moles of each ion.
The density of the solution is $3.107$ g/ml or $3.107$ kg/L.
Thus, $1$ L of the solution will weigh $3.107$ kg.
The molar mass of ammonium chloride is $53.5$ g/mol.
The molality of each ion is $\frac{\text{number of moles of each ion}}{\text{mass of solvent (in kg)}}=\frac{2}{3.107-\frac{2\times 53.5}{1000}}=0.67m$.