(પાણી માટે $K_f = 1.86\,^oC\, kg\, mol^{-1}$ છે)
(મોલર દળ $S = 32\, g\, mol^{-1}, Na = 23\, g\, mol^{-1}$ )
\( = \,\frac{{\Delta {T_f}}}{{{K_f}}}\, = \,\frac{{3.82}}{{1.86}}\, = \,2.054\,mol/1000\,g\) solvent
Molality (theoretical)
\( = \,\frac{{mole\,\,of\,solute}}{{wt.\,of\,solventing\,(g)}} \times 1000\)
\( = \,\frac{{5\,g\,/\,142\,g\,/\,mole}}{x} \times 1000\)
\(N{a_2}S{O_4}\, \to \,2N{a^ + }\, + \,SO_4^{2 - }\)
Moles before dissociation \(1\) \(0\) \(0\)
Moles after dissociation \(1-x\) \(2x\) \(x\)
Von't Hoff Factor \((i) = \,\frac{{Moles\,after\,\,dissociation}}{{Moles\,before\,\,dissociation}}\)
\( = \frac{{(1 - x)\, + \,2x\, + \,x}}{1}\)
\(Na_2SO_4\) is ionised \(81.5\%\) means \(x\,=\,0.815\)
\( = \frac{{(1 - 0.815)\, + \,2 \times 0.815\, + \,0.815}}{1}\)
\(=\,2.63\)
\(i\, = \,\frac{{Observed\,\,molarrity}}{{Calculated\,\,molarity}}\)
\( \Rightarrow \,2.63\, = \,\frac{{2.054}}{{\frac{{0.0352}}{x} \times 1000}}\, = \,45.07\,g\)
આણ્વિય દળ ${K}=39, {Mn}=55, {O}=16]$
બેન્ઝીનના દ્રાવણમાં ઉત્કલન બિંદુમાં વધારો ${ }^{\circ} {C}$માં ${x} \times 10^{-2}$ છે.${x}$નું મૂલ્ય $.....$ છે.(નજીકના પૂર્ણાંકમાં)
$[$ આણ્વિય દળ : ${C}=12.0, {H}=1.0, {O}=16.0]$