Question

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

Answer 1

(A) $M=\frac{\text{percentage by weight}\times 10\times d}{M{w}_2}=\frac{90\times 10\times 1.8}{98}=18M$
$18M\times V_L=3.0M\times 3L\Rightarrow V_L=0.5L=500mL$
$500mL$ of $98$ % $H_{2}S{O}_4$ by mass (density $=1.8gm{L}^{-1})$ is required to prepare $3.0$ L of $3.0$ M $H_{2}S{O}_4$.
(B) Moles of $NaHC{O}_3=\frac{21.0}{84}=0.25mol$
Moles of $HCl=$ Moles of $NaHC{O}_3$
$3M\times V_L=0.25mol;V_L=0.083L=83.3mL$
$83.3mL$ of $3.0M$ $HCl$ should be added to react completely with $21.0g$ of $NaHC{O}_3$.
(C) Moles of $Zn=\frac{6.54}{65.4}=0.1mol$
Moles of $H_{2}S{O}_4=$ moles of $Zn$
$3M\times V_L=0.1,V_L=0.033L=33.3mL$
$33.3mL$ of $3.0M{H}_{2}S{O}_4$ is required to react with $6.54g$ of $Zn$.

(D) Moles of $K_2{C}_2{O}_4\cdot H_{2}O=\frac{92}{184}=0.5mol$
$5mol$ of $C_2{O}_4^{2-}\equiv 2$ mol of $Mn{O}_4^{\ominus }$
$0.5mol$ of $C_2{O}_4^{2-}\equiv 0.2$ mol of $Mn{O}_4^{\ominus }$
$M\times V_L$ of $Mn{O}_4^{\ominus }\equiv 0.2$ mol of $Mn{O}_4^{\ominus }$
$0.5\times V_L=0.2\Rightarrow V_L=0.4L=400mL$
$400mL$ of $0.5MKMn{O}_4$ solution will react completely with $92.0g$ of $K_2{C}_2{O}_4\cdot H_{2}O$.