Three vessels ((A, B, C) ) contain ( H ₂ O ₂ ) solution. In vessel (A, 500 ~mL ) of (10 vol H ₂ O ₂ ) is present. (100 ~mL ) of (30 vol H ₂ O ₂ ) is present in vessel (B ). Vessel (C ) is filled with (250 ~mL ) of (2 M H ₂ O ₂ ). The weight (in ( g ) ) of ( H ₂ O ₂ ) present in these vessels follows the order

Step-by-step solution

For vessel (A ), (10 vol H ₂ O ₂ ) means (1 ~mL ) of ( H ₂ O ₂ ) solution gives (10 mLO ₂ ). ( array cc 2 H ₂ O ₂ & 2 H ₂ O + O ₂ 2 ~mol & 1 ~mol =68 ~g & 22400 ~mL at NTP array ) ( 22400 ~mL ₂ ) is obtained by (68 ~g H ₂ O ₂ ). ( 10 ~mL O ₂ ) is obtained by ( 68 10 22400 =0.03 ~g H ₂ O ₂ ) Vessel (B ), (30 vol H ₂ O ₂ ) means (1 ~mL ) of ( H ₂ O ₂ ) solution gives (30 mLO ₂ ). Means, (22400 ~mL ) of ( O ₂ ) is obtained by (68 ~g H ₂ O ₂ ). ( 30 ~mL O ₂ ) is obtained by ( 68 30 22400 =0.09 ~g H ₂ O ₂ ) Vessel (C ),