{:("Column-I","Column-II"),("A) 1.2g of carbon","P) "4.8xx10^(23)" atoms"),("B) 3g of "CO_(2),"Q) "9.034xx10^(23)" atoms"),("C) 11.2 L "CO_(2)" at S.T.P.","Q) 0.1 moles"),("D) 0.2 moles "NH_(3),"S) "9.64xx10^(24)" electrons"):}

{:("Column-I","Column-II"),("A) 1.2g of carbon","P) "4.8xx10^(23)" ato

1.	{:("Column-I","Column-II"),("A) 1.2g of carbon","P) "4.8xx10^(23)" atoms"),("B) 3g of "CO_(2),"Q) "9.034xx10^(23)" atoms"),("C) 11.2 L "CO_(2)" at S.T.P.","Q) 0.1 moles"),("D) 0.2 moles "NH_(3),"S) "9.64xx10^(24)" electrons"):}
Answer» SOLUTION :A) 1.2g C = 0.1 MOLES B) 32g `O_(2)` implies 1 mole = `6XX10^(23)xx` atoms `=12xx8xx10^(23)e^(-)` C) 11.2L `CO_(2)` `implies (1)/(2)` mole = `(1)/(2)xx6xx10^(23)xx` 3atoms D) 0.2 moles `NH_(3)=0.2xx6xx10^(23)xx4` atoms