Question

$N_2{O}_5+H_{2}O\to 2HN{O}_3$
The concentration of a mixture of $HN{O}_3$ and $N_2{\left(O_5\right)}_{(g)}$ can be expressed similar to oleum. Initially, we have a mixture containing $23g\text{of}HN{O}_3$ and $27g\text{of}{N}_2{\left(O_5\right)}_{(g)}$ . Find the percentage labelling if $100g$ of this mixture is mixed with $4.5g\text{of}{H}_{2}O$ . (Give answer after rounding off to the nearest integer value.)

Answer 1

Answer: 104

(i) 23 g HNO₃
$27\text{g}{\text{ N}}_2{\text{O}}_5\equiv \frac{27\text{g}}{108\text{g}}=\frac{1}{4}$ mole will add $\frac{1}{4}$ mole H₂O
i.e 4.5 g H₂O
50 g mixture needs 4.5 g H₂O
100 g mixture needs 9 g H₂O
% labelling is 109%
(ii) Now in 100 g mixture 4.5 g H₂O added
46 g HNO₃
54 g N₂O₅ (out of this 27 g N₂O₅ will dissolve in 4.5 g) H₂O . plus another 4.5 g H₂O can be added/
For 104.5 g mixture water to be added is 4.5 g
for 100 g mixture water to be added is
$\frac{4\cdot 5}{104\cdot 5}\times 100=4\cdot 3$

$\%$ labelling now $=104.3\simeq 104$