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A gasesous mixture contains three gases N2, O2 and  CO2 with a total number of moles of  10 and total pressure of  10 atm. The partial pressure of  N2 is  3 atm . The pressure is reduced to  9 atm if the gases is passed through alkaline pyrogallol. If  CO2 has molecular weight of   44 g/mol. Then, the weight of  CO2 (in grams) present in the mixture will be:
 


Solution

Given total pressure (P)=10 atm
Given,
total number of moles; nN2+nO2+nCO2=10
The gas O2 is absorbed by pyrogallol. Hence the loss of pressure is due to O2.

10nO2nN2+nO2+nCO2×10=9 atm


Hence the number of moles of O2 is 1.
PN2=3 atm, PO2=1 atm, nN2=3, nO2=1
PN2=xN2×P(total)=nN2nN2+nO2+nCO2×10
=nN210×10nN2=3

 nCO2=10(nN2+nO2)=104=6
Weight of C=6×44=264 g

Chemistry

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