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Mole fraction of K_(2)CO_(3) in a mixture of K_(2)CO_(3) and KHCO_(3) is 0.5. What will be the volume of 0.1 N HCl required to neutralize 1.252 g of the mixture? |
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Answer» Solution :As MOLE fraction of `K_(2)CO_(3)` is 0.5, this mean the mixture contains equal number of moles of `K_(2)CO_(3)` and `KHCO_(3)`. Suppose no. of moles of each is x. `"No. of g eq. in x moles of "K_(2)CO_(3)=2X.` `"No. of g eq. in x moles of "KHCO_(3)=x` `"Total g equivalents of "K_(2)CO_(3) and KHCO_(3)=2x+x=3X` Suppose volume of 0.1 N HCl REQUIRED to neutralize the mixture = V mL `therefore"No. of g equivalents present in V mL of 0.1 HCl "=(0.1 xxV)/(1000)"g eq."` As acids and bases are neutralized in equivalent amounts, `3x=(0.1xxV)/(1000) or V=3xx10^(4)x` `"Molecular mass of "K_(2)CO_(3)=2xx39+12+48=138` `"Molecular mass of "KHCO_(3)=39+1+12+48=100` `therefore x" mole of "K_(2)CO_(3)="138x gandx moles of "KHCO_(3)=100xg` `therefore 138x+100x=1.252 or x=(1.252)/(238)=5.26xx10^(-3)` `therefore V=3xx10^(4)xx5.26xx10^(-3)=157.8mL` |
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