Question

Match the reactions given in column I with molarities given in column II.

Answer 1

(A) millimoles of $Zn=\frac{196.2mg}{65.4mg}=3mmol$

3 mmol of Zn $\equiv$ 2 mmol of $lFe(CN{)}_6{]}^{4-}$

$\equiv \frac{2mmol}{40mL}=0.05M$

0.05 M $K_4\left[Fe\right(CN{)}_6]$ should be used so that 40 mL of a solution titrates 196.2 mg $Zn$ (dissolved) by forming $K_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2$.

(Atomic weight $Zn=65.4gmo{l}^{-1})$
(B) Moles of $BaS{O}_4=\frac{2.33}{233}=0.01mol$

$=0.01molN{a}_{2}S{O}_4$

$M_{N{a}_{2}S{O}_4}=\frac{mmol}{mL}=\frac{0.01\times {10}^3}{40}=0.25M$

The molarity of the $N{a}_{2}S{O}_4$ solution is 0.25 M.
(C) mmol of $T{h}^{4+}=\frac{116mg}{232mg}=0.5mmol$
$=0.5\times 2\times$ mmol of $H_2{C}_2{O}_4$

$M_{H_2{C}_3{O}_4}=\frac{mmol}{mL}=\frac{0.5\times 2}{50}=0.02M$

The molarity of $H_2{C}_2{O}_4$ is 0.02 M.
(D) mmol of $AgN{O}_3=10\times 0.5=5mmol$

$=2\times 5mmol$ of $NaCN$
$M_{NaCN}=\frac{mmol}{mL}=\frac{2\times 5}{100}=0.1M$

The molarity of $NaCN$ is $0.1M$.