Question

What weight of $ N{a_2}C{O_3} $ of $ 85\% $ purity would be required to prepare $ 45.6{\text{ ml}} $ of $ 0.235{\text{ M}} $ $ {H_2}S{O_4} $ ?

Answer 1

Hint: To proceed with this question, we will apply the rule of neutralization reaction. So basically in this question it is asked that what weight of $ N{a_2}C{O_3} $ of $ 85\% $ purity would be required to neutralize $ 45.6{\text{ ml}} $ of $ 0.235{\text{ M}} $ $ {H_2}S{O_4} $ .

Complete answer:
Since, because of the neutralization reaction we can say that:
Milliequivalents of $ N{a_2}C{O_3} $ will be equal to the milliequivalents of the $ {H_2}S{O_4} $ .
Hence , we can write:
Meq of $ N{a_2}C{O_3} $ $ = $ Meq of $ {H_2}S{O_4} $ , For complete neutralization reaction.
Now, Meq of $ N{a_2}C{O_3} $ $ = $ $ 0.235 \times 45.6 $
Also by the formula , we can write the Meq of $ N{a_2}C{O_3} $ in other way as:
 $ \dfrac{w}{{equivalent{\text{ }}weight}} \times 1000 = 45.6 \times 0.235 $
So, let's find out the equivalent weight of $ N{a_2}C{O_3} $ .
Also , note that here we have used $ 1000 $ in the equation because everything is in milliequivalents
So, our equation becomes:
 $ \dfrac{w}{{\dfrac{{106}}{2}}} \times 1000 = 45.6 \times 0.235 $
 $ w = 0.5679{\text{ g}} $
Hence our calculated weight of $ N{a_2}C{O_3} $ is , $ w = 0.5679{\text{ g}} $
Now, moving forward , it is given in the question that $ N{a_2}C{O_3} $ is $ 85\% $ pure.
It means that,
 $ 85 $ gram $ N{a_2}C{O_3} $ will be present in $ 100 $ gram of sample
So, $ 1 $ gram of $ N{a_2}C{O_3} $ will be present in $ = \dfrac{{100}}{{85}} $ gram of sample
And $ 0.5679{\text{ }} $ gram of $ N{a_2}C{O_3} $ will be present in $ = \dfrac{{100}}{{85}} \times 0.5679 $
 $ = 0.668 $ gram
Therefore the correct answer is $ = 0.668 $ gram
So, $ 0.668 $ $ N{a_2}C{O_3} $ of $ 85\% $ purity would be required to prepare $ 45.6{\text{ ml}} $ of $ 0.235{\text{ M}} $ $ {H_2}S{O_4} $ .

Note:
Miliequivalent is one- thousandth of the equivalent of the chemical species , compound , element or radical. An equivalent is the amount of substance that reacts with an arbitrary amount of another substance in a particular chemical reaction. It is a unit of measurement of the compounds that are thoroughly used in the chemistry and the biological sciences. The mass of an equivalent is called its equivalent weight.