Question

When $9.45g$ of $ClC{H}_{2}COOH$ is added to $500mL$ of water, its freezing point drops by $0.5^{\circ }C$ . The dissociation constant of $ClC{H}_{2}COOH$ is $x\times 10^{-3}$ . The value of x is _______. (Rounded off to the nearest integer)
$\left[K_{f\left(H_{2}O\right)}=1.86Kkg{\left(mol\right)}^{-1}\right]$

Answer 1

Answer: 35

Total no. of moles $=C+C\alpha =C(1+\alpha )$
$i=\frac{observedC.P.}{calculatedC.P.}=\frac{C(1+\alpha )}{C}=\left(1+\alpha \right)$
$M.W.=94.5$
$\Delta T_f=i\times k_f\times m\Delta T_f=0.5^{\circ }C$
$i=1+\alpha$
$0.5=\left(1+\alpha \right)\times 1.86\times \frac{\frac{9.45}{94.5}}{\frac{500}{1000}}$
$m=\frac{mole}{k.g(Solvent)}$
$k_t=1.86kkg/mol$
$(1+\alpha )=\frac{2.5}{1.86}$
$\alpha =\frac{0.64}{1.86}=\frac{32}{93}$
$C=\frac{9.45}{94.5\times 0.5}=0.2M$
$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.2\times 1024}{93\times 93\times \frac{61}{93}}$
$K_a=0.0351=35.1\times 10^{-3}$