Question

Match the items given in list I with those in list II.

Answer 1

(A) $2$ mol $N{H}_4^{\oplus }\equiv 1$ mol of $(N{H}_4{)}_{2}C{O}_3=96$ g
$0.4$ mol $N{H}_4^{\oplus }=\frac{96\times 0.4}{2}=19.2$ g
Thus, $19.2$ g of $(N{H}_4{)}_{2}C{O}_3$ (Mw $=96$ g/mol) contains $0.4$ mol $N{H}_4^{\oplus }$.
(B) $8$ mol $H$ atoms $\equiv 1$ mol $(N{H}_4{)}_{2}C{O}_3=96$ g
$1$ mol $H$ atoms $(6.02\times {10}^{23}$ $H$ atoms) $=\frac{96}{8}=12.0$ g
Thus, the mass of $(N{H}_4{)}_{2}C{O}_3$ which contains $6.02\times {10}^{23}$ hydrogen atoms is $12.0$ g.
(C) $\underset{1mol}{(N{H}_4{)}_{2}C{O}_3}+2HCl\to 2N{H}_{4}Cl+\underset{1mol}{C{O}_2}+H_{2}O$
$1$ mol of $C{O}_2\equiv 1$ mol of $(N{H}_4{)}_{2}C{O}_3=96$ g
$3$ mol of $C{O}_2\equiv 96\times 3=288.0$ g
Thus, the mass of $(N{H}_4{)}_{2}C{O}_3$ which will produce $3.0$ mL of $C{O}_2$ when treated with sufficient acid is $288.0$ g.
(D) $100$ mL $\times 0.2M=20$ mmol $(N{H}_4{)}_{2}C{O}_3$
$=20\times {10}^{-3}\times 93=1.92$ g
Thus, the mass of $(N{H}_4{)}_{2}C{O}_3$ required to prepare $100$ mL of $0.2M(N{H}_4{)}_{2}C{O}_3$ solution is $1.92$ g.