{:("Column-I","Column-II"),("A) 22.4 volume "H_(2)O_(2),"P) 11.2 litres"),("B) 32 g oxygen","Q) "9.034xx10^(23)" atoms"),("C) 11.2 L "CO_(2)" at STP","R) 2 mole"L^(-1)),("D) 1 gram equivalent hydrogen at STP","S) "9.64xx10^(24)" electron"):}

{:("Column-I","Column-II"),("A) 22.4 volume "H_(2)O_(2),"P) 11.2 litre

1.	{:("Column-I","Column-II"),("A) 22.4 volume "H_(2)O_(2),"P) 11.2 litres"),("B) 32 g oxygen","Q) "9.034xx10^(23)" atoms"),("C) 11.2 L "CO_(2)" at STP","R) 2 mole"L^(-1)),("D) 1 gram equivalent hydrogen at STP","S) "9.64xx10^(24)" electron"):}
Answer» Solution :A) 1 vol `H_(2)O_(2)=3.03g//L` 22.4bol `H_(2)O_(2)=22.4xx3.03=67.87g//L` `=~~2" moles/L"` B) 32 G oxygen = 1 mole oxygen = `6.023xx10^(23)` MOLECULES `=6.023xx10^(23)xx16=9.64xx10^(24)` ELECTRONS C) 11.2L `CO_(2)` at STP `=(1)/(2)` mole `CO_(2)=(1)/(2)xx6.023xx10^(23)` molecules `=(1)/(2)xx6.023xx10^(23)xx3` ATOM `=9.034xx10^(23)` atom D) 1 gram equivalent HYDROGEN = 1 g hydrogen