Question

The amount of $(N{H}_4{)}_{2}S{O}_4$ having degree of dissocation 75% which should be dissolved in 1500 ml of 1 M $N{H}_{4}OH$ to decrease its degree of dissociation by 200 times, is [$k_b$ of $N{H}_{4}OH$ = $1.8$ $\times {10}^{-5}$]:

• A. 112.1gm
• B. 224.2 gm
• C. 56.0gm
• D. 65.4gm

Answer 1

The correct option is A 112.1gm

To calculate the concentration of ammonium sulfate in the solution, the expression is:
$k_b=\frac{\frac{volumeofN{H}_{4}OH\times c\times degreeofdissociationofN{H}_{4}OH}{200}}{concentrationofN{H}_{4}OH}$
$1.8\times {10}^{-5}=\frac{\frac{1.5c\times 4.24\times {10}^{-3}}{200}}{1}$
Hence the concentration c is $0.57M$.
The number of moles of ammonium sulfate that should be dissolved in 1500 ml of ammonium hydroxide solution is $0.57\times \frac{1500}{1000}=0.885$ moles.
The mass of ammonium sulfate that should be dissolved in 1500 ml of ammonium hydroxide solution is $0.855\times 132=112.1g$.