Get Answers to all your Questions

header-bg qa

A 10mg  efferovescent tablet antaining sodium bicarbanate and oxalic acid releases 0.25 ml of  \mathrm{CO}_2  at T=298.15K and P=1 ban Sf molar volume of \mathrm{CO}_2  is 250 L unders such cendition, what is the percentage of sodium bicarbanates in each tablet

Option: 1

8.4


Option: 2

0.84


Option: 3

16.8


Option: 4

33.6


Answers (1)

best_answer

2 \mathrm{NaHCO}_3+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow 2 \mathrm{CO}_2+\mathrm{Na}_2 \mathrm{C}_4 \mathrm{O}_4+\mathrm{H}_2 \mathrm{O} \\
\mathrm{2~ mol}                    \mathrm{1~ mol}                 \mathrm{2~ mol}

\mathrm{\Rightarrow} In the reaction, number of mole of  \mathrm{\mathrm{CO}_2}  produced.
\mathrm{ n=\frac{P V}{R T}=\frac{1 \text { bar } \times 0.25 \times 10^{-3} \mathrm{~L}}{0.082 \mathrm{Latm} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 298.15 \mathrm{~K}} }
\mathrm{ =1.08 \times 10^{-5} \mathrm{~mol} }

Namber of mole of  \mathrm{ {NaHCO}_3 }\mathrm{ =\frac{Wt. of \mathrm{NaHCO}_3}{ Molecular~ mass ~of~ \mathrm{NaHCO}_3} }

\mathrm{ w_{\mathrm{NaHCO}_3} =1.08 \times 10^{-5} \times 84 \times 10^3 \mathrm{mg} \\ }
\mathrm{ =0.856 \mathrm{mg} \\ }
\mathrm{ \Rightarrow \mathrm{NaHCO}_3 \% =\frac{0.856}{10} \times 100=8.56 \% }

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE