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The relative densit of a mixture of nitrogen and oxygen is 154.4(H=1) and the relative densities of nitrogen and oxygen are 14.0 and 16.0 (H=1) respectively. Calculate the composition of the mixture (i) by volume and (ii) by mass. |
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Answer» Solution :(i) LET 1 mL of the mixture contain x mL of `N_(2)` and (1-x) mL of `O_(2)`. Mass of x mL of `N_(2)=14.0xx x=14x g` (Mass `=dxxF`) Mass of `(1-x)mL` of `O_(2)=16.0xx(1-x)=(16.0-16x)g` Total mass of mixture `=14x+16.0-16x` So, `14x+16.0-16x=14.4xx1=14.4` or `x=0.8`, i.e., 80% by volume Oxygen `=1-x=(1-0.8)=0.2`, i.e. 20% by volume. (ii) Let 1 g of the mixture contain x g of `N_(2)` and (1-x)g of oxygen. Volume of x g of `N_(2)=(x)/(14.0) (V=("mass")/("DENSITY"))` Volume of `(1-x)g` of `O_(2)=((1-x))/(16.0)` Total volume of the mixture `=(x)/(14.0)+((1-x))/(16.0)` So, `(x)/(14.0)+((1-x))/(16.0)=(1)/(14.4)` or `x=0.7778`, i.e., 77.78% by mass Oxygen `=(1-x)=(1-0.7778)=0.2222` i.e., 22.22% by mass. |
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