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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» 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` |
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