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2 g of benzoic acid (C_(6) H_(5) COOH) dissolved in 25 g of benzene shows a depressionc in freezing point equal to 1.62 K. Molal depression constant for benzene is "4.9 K kg mol"^(-1). What is the percentage association of acid if it forms dinner in solution? How many mL of "0.1 M HCl" are required to react completely with 1 g mixture of Na_(2) CO_(3) and NaHCO_(3) containing equimolar amounts of both? |
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Answer» SOLUTION :`"LET MASS of "Na_(2)CO_(3)=xg` `therefore"mass of "NaHCO_(3)=(1-x)g` `therefore"moles of "Na_(2)CO_(3)="moles of "NaHCO_(3)` `(x)/(106)=(1-x)/(84)` `x=0.558g` `"Number of moles of "Na_(2)CO_(3)=(0.558)/(106)="0.00526 mol"` `"Number of moles of "NaHCO_(3)="0.00526 mol"` `Na_(2)CO_(3)+2HCl rarr 2NaCl +H_(2)O +CO_(2)` `NaHCO_(3)+HCl rarr NaCl+H_(2)O+CO_(2)` Total number of moles of HCl required `=(2xx0.00526)+0.00526` `="0.01578 mol"` `therefore"VOLUME 0.1 M HCl"=("Number of moles")/("molarity")` `=(0.01578)/(0.1)=0.1578L` `=157.8mL` |
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