Question

$N_{2}O_{5}+H_{2}O \rightarrow 2HNO_{3}$
The concentration of a mixture of $HNO_{3}$ and $N_{2}\left(O_{5}\right)_{\left(\right. g \left.\right)}$ can be expressed similar to oleum. Initially, we have a mixture containing $23g\text{of}HNO_{3}$ and $27g\text{of}N_{2}\left(O_{5}\right)_{\left(\right. g \left.\right)}$ . 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

(i) 23 g HNO₃
$2 7 \text{g} \text{ N}_{2} \text{O}_{5} \equiv \frac{2 7 \text{g}}{1 0 8 \text{g}} = \frac{1}{4}$ 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
$\frac{4 \cdot 5}{1 0 4 \cdot 5} \times 1 0 0 = 4 \cdot 3$

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