Question

How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer 1

`75% `of soln = 1125
vol of water = `x`
`n_1v_! + n_2v_2 = n_3v_3`
`n_3 = (n_1v_1 + n_2v_2)/(v_1 + v_2)`
`n_1 = 45%, v_1 = 1125`
`n_2 = 0% , v_2= x%`
`n_3 = (1125 xx 45 + 0 xx x)/(1125 + x)`
`30 > n_3 > 25`
`25 < (1125 xx 45)/(1125 + x) < 30`
`25 xx 1/100 < (1125 xx 45/100)/(1125+x) < 30/100`
`25/100(1125 +x) < (1125 xx 45)/100 < 30(1125+x)/100`
`25(1125+x) < 1125 xx 45 < 30(1125 + x)`
`25(1125 + x) < 1125 xx 45 & 1125 xx 45 < 30(1125 + x)`
`25xx 1125 + 25x < 1125 xx 45`
`25x < 1125 xx 45 - 25 xx 1125`
`x < 4/5 xx 1125`
`x < 900`
And, `30(1125+x) > 1125 xx 45`
`30x > 15 xx 1125`
`x > 1125/2`
`x> 562.5l`
`:. x in (502.5, 900)`
Answer