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?

Explanation

`"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`