In order to find the strength of a sample of sulphuric acid, 10 g were dilluted with water and a piece of marble weighing 7 g placed in it. When all action had ceased, the marble was removed, washed, dried and was found to weight 2.2g. What was the percentage strength of sulphuric acid ?

Explanation

Mass of marble taken = 7.0 g.
Mass of marble left unsed = 2.2 g `therefore` Mass of marble reacted = 7.0-2.2 =4.8 g.
The chemical equation involved in the above problem is
`underset(=100 g)underset(40+12+16xx3)(CaCO_(3))+underset(=98 g)underset(2+32+4xx16)(H_(2)SO_(4))rarr CaSO_(4)+H_(2)O+CO_(2)`
Step 1. To calculate the mass of pure `H_(2)SO_(4)` required to react with 4.8 g of marble.
100 g of marble react with `H_(2)SO_(4)=98 g`
`therefore` 4.8 g of marble will react with `H_(2)SO_(4)=(98)/(100)xx4.8g = 4.704 g`.
Step 2. To calculate the strength of sulpuric acid.
10 g of dil. `H_(2)SO_(4)` contain pure `H_(2)SO_(4)=4.704 g`.
`therefore` 100 g of dil. `H_(2)SO_(4)` will contain pure `H_(2)SO_(4)=(4.704)/(10)xx100g=47.04 g`
Thus, the percentage stremgth of sulphuric acid = 47.04% .