`N_2O_5 + H_2O rarr 2HNO_3`
The concentration of a mixture of `HNO_3` and `N_2O_(5(g))` can be expressed similar to oleum. Initially we have a mixture containing 23g of `HNO_3` and 27g of `N_2O_(5(g))` . Find the percentage labelling if 100g of this mixture is mixed with 4.5g of `H_2O`.

Answer

104.3

Explanation

(i) 23 g `HNO_3`
`27g N_(2)O_5 -= (27g)/(108g) = 1/4` mole will add `1/4` mole `H_2O`
İ.e 4.5 g `H_2O`
50 g mixture needs 4.5 g `H_2O`
100 g mixture needs 9 g `H_2O`
% labelling is 109%
(ii) Now in 100 g mixture 4.5 g `H_2O` added
`46 g HNO_3`
54 g `N_2O_5` (out of this 27 g `N_2O_5` will dissolve in 4.5 g) `H_2O` plus another 4.5 g `H_2O` 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 `(4.5)/(104.5)xx100 =4.3`
% labelling now = 104.3