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  • A 10 mg effer0scent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of <img alt="\mathrm{CO_{2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BCO_%7B2

A 10 mg effer0scent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of \mathrm{CO_{2}}  at T=298.15 K and P=1 bar molar volume of \mathrm{CO_{2}} is 250 under such condition,what is the percentage of sodium bicarbonate in each tablet ___

Option: 1

8.4


Option: 2

0.84


Option: 3

16.8


Option: 4

33.6


Answers (1)

best_answer

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

In the reaction , number of mole of \mathrm{ \mathrm{CO}_2 } produced .

 

\mathrm{n=\frac{P V}{R T}= \frac{1 \text { boro } \times 0.25 \times 10^{-3} \mathrm{~L}}{0.082 \mathrm{Latm} \mathrm{\textrm {k } ^ { - 1 } \mathrm { mol } ^ { - 1 } \times 2 9 8 . 1 5 \mathrm { K }}}}

                   \mathrm{=1.08 \times 10^{-5} \mathrm{~mol}}

Number of mole of \mathrm{\mathrm{NaHCO}_3=\frac{\text { Wt. of } \mathrm{NaHCO}_3}{\text { Molecular man of } \mathrm{NaHCO}_3}}

\mathrm{\begin{aligned} \mathrm{NaHCO}_3 & =1.08 \times 10^{-5} \times 84 \times 10^3 \mathrm{mg} \\ & =0.856 \mathrm{mg} \end{aligned}}

\mathrm{\Rightarrow \mathrm{NaHCO}_1 \%=\frac{0.856}{10} \times 100=8.56 \%}

Posted by

Divya Prakash Singh

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