Exam

JIPMER

Year

2016

Subject

Chemistry

Chapter

Some Basic Concepts of Chemistry

JIPMER 2016 · Chemistry · Some Basic Concepts of Chemistry

Solid NaHCO_3 will be neutralised by 40.0 mL of 0.1 M H_2 SO_4...

Solid $\mathrm{NaHCO}_3$ will be neutralised by 40.0 mL of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution. What would be the weight of solid $\mathrm{NaHCO}_3$ in gram?

Step-by-step solution

The required neutralisation reaction is, $2 \mathrm{NaHCO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{CO}_2$ $m$-moles of $\mathrm{H}_2 \mathrm{SO}_4=M \times V_{\mathrm{mL}}=40.0 \times 0.1=4 \mathrm{~m}$-mole or m-moles of $\mathrm{NaHCO}_3$ neutralised $=4 \times 2=8 \mathrm{~m}$-mole But, m-mole =$\frac{w}{m}$ \times 1000 $$\begin{aligned} 8 & =\frac{w}{84} \times 1000 \\ \Rightarrow \quad w & =\frac{84 \times 8}{1000} \Rightarrow w=0.672 \mathrm{~g}\end{aligned}$$