1.

Commericial sample of H_(2)O_(2)is labeled as 10 V. Its % strength is nearly

Answer»

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Solution :10 V (10 volume) `H_(2)O_(2)` means that 1 mL of `H_(2)O_(2)` solution will produce 10 mL of `O_(2)` at STP.
Now, decomposition of `H_(2)O_(2)` is as below:
`underset((2xx34)g)(2H_(2)O_(2))rarr 2H_(2)O+underset("22400 mL (at STP)")(O_(2))`
`=68g`
`THEREFORE"22400 mL of "O_(2)" is OBTAINED from 68 g of "H_(2)O_(2)`
`therefore"10 mL of "O_(2)" is obtained from "(68)/(22400)xx"10 g of "H_(2)O_(2)`
`=0.03036g" of "H_(2)O_(2)`
but this 10 mL of `O_(2)` corresponds to 0.03 g of `H_(2)O_(2)` solution APPROXIMATELY.
Therefore, 1 mL of `H_(2)O_(2)` solution CONTAINS 0.03 g of `H_(2)O_(2)`
`therefore"100 mL of "H_(2)O_(2)" solution contains 3 g of "H_(2)O_(2)`
`therefore"% strength is 3."`


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