Half litre each of three samples of H-2-O-2- labelled 10 volume- 15 volume and 20 volume are mixed and then diluted with equal volume of water- The relative strength of resultant H-2-O-2- solution is-2-55-07-510

Question

Toppr Admin · Accepted Answer

Volume strength of H2O2-5-6-xD7-N -xA0- -xA0- -N- Normality-Normality of 10 vol of-xA0-H2O2-105-6NNormality of 15 vol of-xA0-H2O2-155-6N

Half litre each of three samples of $H_{2} O_{2}$ labelled $10$ volume, $15$ volume and $20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $H_{2} O_{2}$ solution is:
$2.5$
$5.0$
$7.5$
$10$

Volume strength of $H_{2} O_{2} = 5.6 \times N$ $(N =$ Normality $)$

Normality of $10$ vol of $H_{2} O_{2} = \frac{10}{5.6} N$

Normality of $15$ vol of $H_{2} O_{2} = \frac{15}{5.6} N$

Half litre each of three samples of H2O2 labelled 10 volume, 15 volume and 20 volume are mixed and then diluted with equal volume of water. The relative strength of resultant H2O2 solution is:2.55.07.510

Volume strength of H2O2=5.6×N (N= Normality)Normality of 10 vol of H2O2=105.6NNormality of 15 vol of H2O2=155.6N

Half litre each of three samples of $H_{2} O_{2}$ labelled $10$ volume, $15$ volume and $20$ volume are mixed and then diluted with equal volume of water. The relative strength of resultant $H_{2} O_{2}$ solution is:
$2.5$
$5.0$
$7.5$
$10$

Volume strength of $H_{2} O_{2} = 5.6 \times N$ $(N =$ Normality $)$

Normality of $10$ vol of $H_{2} O_{2} = \frac{10}{5.6} N$

Normality of $15$ vol of $H_{2} O_{2} = \frac{15}{5.6} N$