Question

(a) $2g$ of a metal carbonate were dissolved in $50mL$ of $NHCl$. $100mL$ of $0.1NNaOH$ were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate.
(b) How much water should be added to $75mL$ of $3NHCl$ to make it a normal solution?

Answer 1

A) we have,

$MC{O}_3+2HCl\to MC{l}_2+C{O}_2+H_{2}O$

we have, 50 ml of 1 N HCl

no . of moles of HCl = $\frac{1\ast 50}{1000}$ = 0.05 mole

100 ml of 0.1 N NaOH , have no of moles = 0.01mole

moles of HCl left after neutralisation with NaOH = 0.05 - 0.01 = 0.04 mole

the remaining moles of HCl is reacting with metal carbonate

2 mole of HCl is reacting with 1 mole of metal carbonate

so, 0.04 mole of HCl is reacting with metal carbonate = $\frac{0.04}{2}$ = 0.02 moles

moles of metal carbonate = 0.02 = $\frac{givenweight}{molecularweight}$

we get, molecular weight = 100g

B) we have, $N_1{V}_1=N_2{V}_2$

where, $V_2$ is final volume

so we get, $3.\frac{75}{1000}=V_2\ast 0.1$

$V_2$ = 225 ml

amount of water added = 225 - 75 = 145 ml