1.

Calculate the mass % of `Na_(2)CO_(3)` in a mixture having mass 206 gm which produced 24 litre of `CO_(2)` at 1 atm pressure & 300 K wich axcess of HCl. [R=0.08 atm lit/mok K]A. 0.485B. 0.515C. 0.4D. 0.6

Answer» Correct Answer - 2
`underset("1 mole")(Na_(2)CO_(3))+2HCl to 2NaCl+H_(2)O(l)+underset("1 mole ")(CO_(2)(g))`
`=1xx106=106 gm`
PV=nRT
`n=(1xx24)/(0.8xx300)=24/24=1 ` mole
% of `Na_(2)CO_(3)=106/206xx100=51.5%`


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