Question Number 227919 by Spillover last updated on 01/Mar/26
$$\mathrm{1}{gm}\:{of}\:{KCl}\:{was}\:{dissolved}\:{in}\:\mathrm{130}{gm} \\ $$$${of}\:{water}\:{rising}\:{its}\:{boiling}\:{point} \\ $$$${by}\:\mathrm{0}.\mathrm{1}^{\:°} {C}.{calculate}\:{the}\:{degree}\:{of}\: \\ $$$${dissociation}\:{of}\:{salt}\:{expressed}\:{as} \\ $$$${percentage}\:{Kb}=\mathrm{0}.\mathrm{52}{C}/{mole}/{kg} \\ $$
Answered by TonyCWX last updated on 02/Mar/26
$$\mathrm{Mass}\:\mathrm{of}\:\mathrm{KCl}\:=\:\mathrm{1}\:\mathrm{g} \\ $$$$\mathrm{Molar}\:\mathrm{Mass}\:\mathrm{of}\:\mathrm{KCl}\:=\:\mathrm{39}.\mathrm{0983}\:+\:\mathrm{35}.\mathrm{453}\:=\:\mathrm{74}.\mathrm{5513}\:\mathrm{g}/\mathrm{mol} \\ $$$$\mathrm{Mass}\:\mathrm{of}\:\mathrm{Solvent}\:=\:\mathrm{130}\:\mathrm{g}\:=\:\mathrm{0}.\mathrm{13}\:\mathrm{kg} \\ $$$$\mathrm{Molarity}\:\mathrm{of}\:\mathrm{Solution},\:{m}\:=\:\frac{\mathrm{1}}{\mathrm{74}.\mathrm{5513}×\mathrm{0}.\mathrm{13}}\:\approx\:\mathrm{0}.\mathrm{1031814025}\:\mathrm{mol}/\mathrm{kg} \\ $$$$ \\ $$$$\mathrm{Van}'\mathrm{t}\:\mathrm{Hoff}\:\mathrm{Factor},\:{i}\:=\:\frac{\Delta{T}_{{b}} }{{K}_{{b}} ×{m}}\:=\:\frac{\mathrm{0}.\mathrm{1}}{\mathrm{0}.\mathrm{52}×\mathrm{0}.\mathrm{1031814025}}\:\approx\:\mathrm{1}.\mathrm{8637825002} \\ $$$$ \\ $$$$\mathrm{Degree}\:\mathrm{of}\:\mathrm{Dissociation},\:\alpha\:=\:\frac{{i}−\mathrm{1}}{{n}−\mathrm{1}}\:=\:\frac{\mathrm{1}.\mathrm{8637825002}−\mathrm{1}}{\mathrm{2}−\mathrm{1}}\:\approx\:\mathrm{0}.\mathrm{8637825002} \\ $$$$ \\ $$$$\mathrm{Percentage}\:=\:\mathrm{86}.\mathrm{37825002\%}\:\approx\:\mathrm{86}.\mathrm{38\%} \\ $$
Commented by Spillover last updated on 02/Mar/26
$${correct} \\ $$
Answered by Spillover last updated on 02/Mar/26
Commented by Spillover last updated on 02/Mar/26
$${i}={vant}\:{Hoff}'{s}\:{factor} \\ $$
Commented by Spillover last updated on 02/Mar/26
$${Dissociation}\:{of}\:{KCl} \\ $$$${KCl}\rightarrow{K}^{\oplus} +{Cl}^{\circleddash} \:\:\: \\ $$$${n}=\mathrm{2} \\ $$
Commented by Spillover last updated on 02/Mar/26
$$\bigtriangleup{T}_{{Cal}} =\mathrm{0}.\mathrm{054}^{°} {C} \\ $$$$\bigtriangleup{T}_{{Observed}} =\mathrm{0}.\mathrm{1}^{°} {C} \\ $$$${i}=\mathrm{1}.\mathrm{862} \\ $$