Calculate the percentage of anhydrous sodium carbonate in the sample if a solution containing 2.5 g sodium carbonate sample is neutralized completely by 55 cc of N/2 HCL.

Solution:

Given,

Weight of Sodium Carbonate= 2.5gm

Volume of HCL(V) = 55ml

Normality of HCL (N) = $\frac{1}{2}$N=0.5N

So, Purity Percentage = $\frac{Weight\;of\; HCL}{Weight \;of\; Na_2CO_3}$

Now, 

We Know that:

$\frac{W}{E}$ = $\frac{NV}{1000}$

Or, $\frac{W}{53}$=$\frac{(0.5)(55)}{1000}$

On Solving, Above equation,

W = 1.4575gm

Thus, Purity Percentage = $\frac{Weight\;of\; Na_2CO_3\;under\; reaction}{Weight \;of\; Na_2CO_3}$ × 100%

=$\frac{1.4575}{2.5}$× 100%

=58.3%

Thus, the percentage of anhydrous sodium carbonate required is 58.3.