Question

Find the amount of $98\% $ pure $N{a_2}C{O_3}$ required to prepare $5$ litres of $2N$ solution.
A. $54.8g$ impure $N{a_2}C{O_3}$
B. $540.8g$ impure $N{a_2}C{O_3}$
C. $40.8g$ impure $N{a_2}C{O_3}$
D. $340.8g$ impure $N{a_2}C{O_3}$

Answer 1

Hint: We will approach this problem with the help of some important formulae and these formulae are (I) no of equivalents = N $\times$ V, where N is normality and V is volume, (ii) no of equivalent=weight /equivalent weight.

Complete answer:
So now we will calculate equivalent weight of $N{a_2}C{O_3}$, we know that molecular weight of $N{a_2}C{O_3}$ is $2 \times 23 + 12 + 3 \times 16 = 106$ (molar mass of sodium ,carbon and oxygen atoms are $23$, $12$ and $16$) and n-factor of Sodium carbonate salt is 2 as there are two positive charges present in electropositive cation of sodium salt. Now equivalent weight = molecular weight /n-factor so equivalent weight of $N{a_2}C{O_3}$= $\dfrac{{106}}{2}$$ = 53$
No of equivalent of $N{a_2}C{O_3}$=$5 \times 2 = 10$
We also know that no of equivalent of Sodium carbonate ( $N{a_2}C{O_3}$)= weight /equivalent weight
No of equivalent $ \Rightarrow \dfrac{{weight}}{{53}}$
$ \Rightarrow 10 = \dfrac{{weight}}{{53}}$
$ \Rightarrow weight = 530 g$

But here in the problem it is mention that we have to find out the amount of $98\% $ pure $N{a_2}C{O_3}$ so we know that $98g$ of the $N{a_2}C{O_3}$ is present in the $100g$ sample. So the amount of $N{a_2}C{O_3}$ present in $530g$ of the sample is equal to $\dfrac{{530 \times 100}}{{98}} = 540.8g$. So option B is correct, that is $540.8g$ impure $N{a_2}C{O_3}$.

Note: We have solved this problem by finding a number of equivalents. Once number of equivalent is known then we have calculated the weight of substance ($N{a_2}C{O_3}$) but the result we got it is when we have assumed that $N{a_2}C{O_3}$ is $100\% $ pure. Now as it is asked in the question to calculate the amount of the $98\% $ pure $N{a_2}C{O_3}$ so we have found out the amount which is asked. So the amount of $98\% $ pure $N{a_2}C{O_3}$ required to prepare $5$ litres of $2N$ solution is $540.8g$ impure sodium carbonate.