(a) What is van't Hoff factor? What types of values it can have if the solute molecules undergo
(i) Dissociation ?
(ii) Association ?
(b) How many mL of a 0.1 M HCl solution are required to react completely with 1 g of a mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` containing equimolar amounts of both ?
(Molar mass : `Na_(2)CO_(3)=106g, NaHCO_(3)=84 g`)

Explanation

The van't Hoff factor I (named after J.H. van't Hoff) is a measure of the effect of a solution upon colligative properties such as osmotic pressure, relative lowering in vapour pressure elevation of boiling point and freezing point depression. The van't Hoff factor is the ratio between the actual concentration of particles produced when the subtance is dissolved , and the concentration of a substance as calculated from its mass.
Dissociated solutes: The degree of dissociation is the fraction of the original solute molecules that have dissociated . It is usually indicated by the Greek symbol `alpha`. There is a simple relationship between this parameter and the van't Hoff factor. If a fraction `alpha` . There is a simple relationship between this parameter and the van't Hoff factor. If a fraction `alpha` of the solute dissociates into n ions, then
`" "i=alpham+(1-alpha)=1+alpha(n-1)`
For example, the dissociation:
`" "KCl hArr K^(+)+Cl^(-)`
yields n=2 particles, so that `i=1+alpha`
Associated solutes : Similarly, if a fractiion `alpha` of n moles of solute associate to form one mole of an n-mer (dimer , trimer, etc.) , then
`" "i=1-(1-(1)/(n))alpha`
For the dimerisation of acetic acid in benzene : `2CH_(3)COOH hArr (CH_(3)COOH)_(2)`
2 moles of acetic acid assoicate to form 1 mole of dimer, so that
`" "i=1-(1-(1)/(n))alpha=1-(alpha)/(2)`
(b) Let the amount of `Na_(2)CO_(3)` be X
Let teh amount of `Na_(2)HCO_(3)` be 1-X
Since no. of moles of both are equal
`(X)/(N(Na_(2)CO_(3)))=(1-X)/(M(NaHCO_(3)))`
`(X)/(106)=(1-X)/(84)`
84X=106-106X
X=0.5578
`nNa_(2)CO_(3)=0.5578//106=0.00526`
`nNaHCO_(3)=0.00526`
`Na_(2)CO_(3) + 2HCl to 2NaCl + CO_(2) + H_(2)`
`NaHCO_(3) + HCl to NaCl + CO_(2) + H_(2)O`
`M_(1)V_(1) = 2M_(2)V_(2)+ M_(3)V_(3)`
`0.1 xx V_(1)=2xx0.00526 + 0.00526`
`V_(1)=(0.01578)/(0.1)=0.1578 L`
`V=157.8 ml.`