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10 mL of a given solution of H_(2)O_(2) contains 0.91 g of H_(2)O_(2) . Express its strength in volume. |
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Answer» Solution :68 g of `H_(2)O_(2)` produce `O_(2)=22400` ML at NTP `therefore 0.91 g ` of `H_(2)O_(2)` will produce `O_(2)=(22400xx0.91)/(68)~= 300` mL at NTP `therefore ` Volume STRENGTH `~=(300)/(10)~=30` |
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