InterviewSolution
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InterviewSolution
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\left. \begin{array} { l l } { ( A ) 22 : 3 : 7 } & { ( B ) 0.5 : 3 : 7 } \\ { ( C ) 1 : 3 : 1 } & { ( D ) 1 : 3 : 0.5 } \end{array} \right. |
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Answer» Number of moles of CO2 = given weight/molecular weight= 22/44= ½Number of moles per V liters of volume = ½V2:Number of moles of H2 = given weight/molecular weight= 3/2Number of moles per V liters of volume = 3/2V3:Number of moles of N2 = given weight/molecular weight= 7/28= ¼Number of moles per V liters of volume = ¼V Total number of moles = nCO2+ nH2+ nN2= ½V + 3/2V +¼V= 9/4V Ratio of number of moles in comparison with total number of moles:= (½V)/(9/4V) : (3/2V)/(9/4V) : (¼V)/(9/4V)= 2/9 : 6/9 : 1/9 Ratio of active masses of gases = 2 : 6 : 1 |
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