Question

100 " mL of " each thhree samples of `H_2O_2` labelled 2.8 vol 5.6vol, and 22.4 vol are mixed and then diluted with an equal volume of water. Calculate the volume strength of the resultant `H_2O_2` solution.

Answer 1

Volume strength of `H_2O_2=5.6xxN`
`N_1 of 2.8 vol H_2O_2=(2.8)/(5.6)=0.5N`
`N_2 of 5.6 vol H_2O_2=(5.6)/(5.6)=1N`
`N_3 of 22.4 vol H_2O_2=(22.4)/(5.6)=4N`
`N_1V_1+N_2V_2+N_3V_3=N_4V_4(V_4=300+300=600mL)`
`0.5xx100+1xx100+4xx100=N_4xx600`
`N_4=(50+100+400)/(600)=(550)/(600)=0.91N`
Volume strength`=5.6xx0.91=5.09vol`
Alternatively:
(Volume strength`)_1+`(Volume strength`)_2`
`+("Volume strengt"h)_3`
`=(2.8+5.6+22.4)/(6)=(30.8)/(6)=5.1vol`