Calculate the number of moles in : (i) 392 grams of sulphuric acid (ii) 44.8 litres of carbon dioxide at STP (iii) 6.022xx10^(23) molecules of oxygen (iv) 9.0 grams of aluminium (v) 1 metric ton of iron (1 metric ton =10^(3)kg) (vi) 7.9 mg of Ca (vii) 65 mu of carbon.

Calculate the number of moles in : (i) 392 grams of sulphuric acid (ii

1.	Calculate the number of moles in : (i) 392 grams of sulphuric acid (ii) 44.8 litres of carbon dioxide at STP (iii) 6.022xx10^(23) molecules of oxygen (iv) 9.0 grams of aluminium (v) 1 metric ton of iron (1 metric ton =10^(3)kg) (vi) 7.9 mg of Ca (vii) 65 mu of carbon.
Answer» Solution :(i) `"1 MOLE of "H_(2)SO_(4)=98g.""(because" Molecular mass of "H_(2)SO_(4)=2xx1+32+4xx16=98u)` Thus, 98 g of `H_(2)SO_(4)`=1 mole of `H_(2)SO_(4)""therefore"392 g of "H_(2)SO_(4)=(1)/(98)xx392="4 moles of "H_(2)SO_(4)`. (ii) 1 mole of `CO_(2)=22.4` litres at STP, i.e., 22.4 litres of `CO_(2)` at STP = 1 mole `therefore 44.8 ` litres of `CO_(2)` at STP = `(1)/(22.4)xx44.8="2 moles "CO_(2)` (iii) 1 mole of `O_(2)` molecules =`6.022xx10^(22)" molecules"` `6.022xx10^(23) " molecules"=" 1 mole of OXYGEN molecules."` (iv) 1 mole of Al = 27 g of Al `""(because"Atomic mass of aluminimum"=27u)` `" i.e. 27 g of aluminium = 1 mole of Al"therefore"9 g of aluminium "=(1)/(27)xx9=0.33" mole of Al"` (v) `"1 metric ton of Fe"=10^(3)kg=10^(6)g` `"1 mole of Fe = 56 g of Fe"therefore""10^(6)g" of "Fe=(10^(6))/(56)" moles"=1.786xx10^(4)"moles".` (vi) 7.9 mg of Ca `=7.9xx10^(-3)" g of Ca"=(7.9xx10^(-3))/(40)"mol"=1.975xx10^(-4)"mol(At. mass of Ca = 40 U)"` (VII) `65.5mu " g of C"=65.5xx10^(-6)g" g of C"=(65.5xx10^(-6))/(12)"mol"=5.458xx10^(-6)"mol"`