1.

Which of the following has maximum mass ? (i) 0.1 mole of KNO_(3) (ii) 1 xx 10^(23) molecules of CH_(4) (iii) 112 cm^(3) of hydrogen at S.T.P ?

Answer»


Solution :(i) Formula mass of `KNO_(3)=1 xx 39 + 1 xx 14 + 3 xx 16 = 101`
1 mole of `KNO_(3)=101 g`
0.1 mole of `KNO_(3)=(("0.1 MOL"))/(("1.0 mol"))xx(101g)=10.1g`
(ii) `6.022 xx 10^(23)` molecules of `CH_(4)=16 g`
`1 xx 10^(23)` molecules of `CH_(4)=((1xx10^(23)"molecules"))/((6.022xx10^(23)"molecules"))xx(16g)=2.657g`
(iii) `22400 cm^(3)` of `H_(2)` at S.T.P = 2g
`112 cm^(3)` of `H_(2)` at S.T.P `= (("112 cm"^(3)))/(("22400 cm"^(3)))xx2g=0.01g`
Thus, 0.1 mole of `KNO_(3)` has MAXIMUM mass.


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