Analytical Chemistry Question 219
Question: 25 ml of a solution of $ Na_2CO_3 $ having a specific gravity of 1.25 required 32.9 ml of a solution of HCl containing 109.5 grams of the acid per litre for complete neutralization. Calculate the volume of $ 0.84NH_2SO_4 $ that will be completely neutralized by 125 grams of the $ Na_2CO_3 $ solution
[UPSEAT 2001]
Options:
A) 460 ml
B) 540 ml
C) 480 ml
D) 470 ml
Show Answer
Answer:
Correct Answer: D
Solution:
$ N_1V_1=N_2V_2 $
$ N\times 25=\frac{109.5\times 32.9}{36.5} $ Þ $ N=\frac{109.5\times 32.9}{36.5\times 25} $
$ N_3V_3=N_4V_4 $ ( $ V_3=\frac{m}{d} $ , $ V_3=\frac{125}{1.25} $ ) $ \frac{109.5\times 32.9}{36.5\times 25}\times 100=0.84\times V $ Þ $ V=470ml $
BETA
