The strengths of three `H_(2)O_(2)` solutions are 10,15 and 20 volume respectively. 0.5L of each of these solutions are mixed and equal amount of water is added to it. Determine the volume strength of the mixed solution.

Explanation

Volume strength = `5.6xx`normality
`therefore` Normality of the first solution, `N_(1)` = 10/5.6, normality of the second solutions are `V_(1), V_(2), V_(3) and V_(R )` respectively and if the normality of the mixed solution is `N_(R )`, then
`V_(1) V_(1) + N_(2)V_(2) + N_(3)V_(3) = N_(R )V_(I)`
or, ` (10)/(5.6) xx (1)/(2) + (15)/(5.6) xx (1)/(2) + (20)/(5.6) xx (1)/(2) = N_(R ) xx 3`
or, ` N_(R ) = ((5 + 7.5 + 10))/(5.6 xx 3 ) = 1.339`
Therefore, the volume strength of the mixed solution = `N_(R ) xx 5.6 = 1.339 xx 5.6 = 7.5 `volume.