The volume of 20-V-H-2-O-2- solution- required to prepare 5 litres of oxygen STP is-125 ml250 ml50 ml100 ml

Question

Toppr Admin · Accepted Answer

20 ml of O2 is prepared by 1 ml of H2O2 at STP-1 ml of-xA0-O2 is prepared by -120 ml of H2O2-5000 ml of-xA0-O2 is prepared by - 120-xD7-5000-250 ml-xA0-of H2O2-Hence- option B is correct-

The volume of $20 V - H_{2} O_{2}$ solution, required to prepare $5$ litres of oxygen at STP is:
$125$ ml
$250$ ml
$50$ ml
$100$ ml

$20$ ml of $O_{2}$ is prepared by $1$ ml of $H_{2} O_{2}$ at STP.

$1$ ml of $O_{2}$ is prepared by $= \frac{1}{20}$ ml of $H_{2} O_{2}$ .

$5000$ ml of $O_{2}$ is prepared by = $\frac{1}{20} \times 5000 = 250$ ml of $H_{2} O_{2}$ .

Hence, option B is correct.

The volume of 20V−H2O2 solution, required to prepare 5 litres of oxygen at STP is:125 ml250 ml50 ml100 ml

20 ml of O2 is prepared by 1 ml of H2O2 at STP.1 ml of O2 is prepared by =120 ml of H2O2.5000 ml of O2 is prepared by = 120×5000=250 ml of H2O2.Hence, option B is correct.

The volume of $20 V - H_{2} O_{2}$ solution, required to prepare $5$ litres of oxygen at STP is:
$125$ ml
$250$ ml
$50$ ml
$100$ ml

$20$ ml of $O_{2}$ is prepared by $1$ ml of $H_{2} O_{2}$ at STP.

$1$ ml of $O_{2}$ is prepared by $= \frac{1}{20}$ ml of $H_{2} O_{2}$ .

$5000$ ml of $O_{2}$ is prepared by = $\frac{1}{20} \times 5000 = 250$ ml of $H_{2} O_{2}$ .

Hence, option B is correct.