The volume of oxgen liberated from 0-96 g of H 2 O 2 at STP is

No.of moles of H2O2 is= 0.96/68=0.014 (68=molecular mass of hydrogen peroxide) H2O2=== gt;H2 + O2 hence volume of O2 evolved =22.4L*0.014=0.3136L=313.6 mL (22 ...

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Standard XII

Chemistry

Abnormal Colligative Properties

The volume of...

Question

The volume of oxgen liberated from 0.96g of H₂O₂at STP is

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Solution

no.of moles of H2O2 is= 0.96/68=0.014 (68=molecular mass of hydrogen peroxide) H2O2===>H2 + O2 hence volume of O2 evolved =22.4L*0.014=0.3136L=313.6 mL (22.4L=volume of 1 mole of ideal gas)

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The volume of oxgen liberated from 0.96g of H2O2 at STP is

no.of moles of H2O2 is= 0.96/68=0.014 (68=molecular mass of hydrogen peroxide) H2O2===>H2 + O2 hence volume of O2 evolved =22.4L*0.014=0.3136L=313.6 mL (22.4L=volume of 1 mole of ideal gas)

The volume of oxgen liberated from 0.96g of H₂O₂at STP is