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Question Number 227919 by Spillover last updated on 01/Mar/26

1gm of KCl was dissolved in 130gm of water rising its boiling point by 0.1^( °) C.calculate the degree of dissociation of salt expressed as percentage Kb=0.52C/mole/kg

$$\mathrm{1}{gm}\:{of}\:{KCl}\:{was}\:{dissolved}\:{in}\:\mathrm{130}{gm} \\ $$$${of}\:{water}\:{rising}\:{its}\:{boiling}\:{point} \\ $$$${by}\:\mathrm{0}.\mathrm{1}^{\:°} {C}.{calculate}\:{the}\:{degree}\:{of}\: \\ $$$${dissociation}\:{of}\:{salt}\:{expressed}\:{as} \\ $$$${percentage}\:{Kb}=\mathrm{0}.\mathrm{52}{C}/{mole}/{kg} \\ $$

Answered by TonyCWX last updated on 02/Mar/26

Mass of KCl = 1 g Molar Mass of KCl = 39.0983 + 35.453 = 74.5513 g/mol Mass of Solvent = 130 g = 0.13 kg Molarity of Solution, m = (1/(74.5513×0.13)) ≈ 0.1031814025 mol/kg Van′t Hoff Factor, i = ((ΔT_b )/(K_b ×m)) = ((0.1)/(0.52×0.1031814025)) ≈ 1.8637825002 Degree of Dissociation, α = ((i−1)/(n−1)) = ((1.8637825002−1)/(2−1)) ≈ 0.8637825002 Percentage = 86.37825002% ≈ 86.38%

$$\mathrm{Mass}\:\mathrm{of}\:\mathrm{KCl}\:=\:\mathrm{1}\:\mathrm{g} \\ $$$$\mathrm{Molar}\:\mathrm{Mass}\:\mathrm{of}\:\mathrm{KCl}\:=\:\mathrm{39}.\mathrm{0983}\:+\:\mathrm{35}.\mathrm{453}\:=\:\mathrm{74}.\mathrm{5513}\:\mathrm{g}/\mathrm{mol} \\ $$$$\mathrm{Mass}\:\mathrm{of}\:\mathrm{Solvent}\:=\:\mathrm{130}\:\mathrm{g}\:=\:\mathrm{0}.\mathrm{13}\:\mathrm{kg} \\ $$$$\mathrm{Molarity}\:\mathrm{of}\:\mathrm{Solution},\:{m}\:=\:\frac{\mathrm{1}}{\mathrm{74}.\mathrm{5513}×\mathrm{0}.\mathrm{13}}\:\approx\:\mathrm{0}.\mathrm{1031814025}\:\mathrm{mol}/\mathrm{kg} \\ $$$$ \\ $$$$\mathrm{Van}'\mathrm{t}\:\mathrm{Hoff}\:\mathrm{Factor},\:{i}\:=\:\frac{\Delta{T}_{{b}} }{{K}_{{b}} ×{m}}\:=\:\frac{\mathrm{0}.\mathrm{1}}{\mathrm{0}.\mathrm{52}×\mathrm{0}.\mathrm{1031814025}}\:\approx\:\mathrm{1}.\mathrm{8637825002} \\ $$$$ \\ $$$$\mathrm{Degree}\:\mathrm{of}\:\mathrm{Dissociation},\:\alpha\:=\:\frac{{i}−\mathrm{1}}{{n}−\mathrm{1}}\:=\:\frac{\mathrm{1}.\mathrm{8637825002}−\mathrm{1}}{\mathrm{2}−\mathrm{1}}\:\approx\:\mathrm{0}.\mathrm{8637825002} \\ $$$$ \\ $$$$\mathrm{Percentage}\:=\:\mathrm{86}.\mathrm{37825002\%}\:\approx\:\mathrm{86}.\mathrm{38\%} \\ $$

Commented by Spillover last updated on 02/Mar/26

correct

$${correct} \\ $$

Answered by Spillover last updated on 02/Mar/26

Commented by Spillover last updated on 02/Mar/26

i=vant Hoff′s factor

$${i}={vant}\:{Hoff}'{s}\:{factor} \\ $$

Commented by Spillover last updated on 02/Mar/26

Dissociation of KCl KCl→K^⊕ +Cl^⊝ n=2

$${Dissociation}\:{of}\:{KCl} \\ $$$${KCl}\rightarrow{K}^{\oplus} +{Cl}^{\circleddash} \:\:\: \\ $$$${n}=\mathrm{2} \\ $$

Commented by Spillover last updated on 02/Mar/26

△T_(Cal) =0.054^° C △T_(Observed) =0.1^° C i=1.862

$$\bigtriangleup{T}_{{Cal}} =\mathrm{0}.\mathrm{054}^{°} {C} \\ $$$$\bigtriangleup{T}_{{Observed}} =\mathrm{0}.\mathrm{1}^{°} {C} \\ $$$${i}=\mathrm{1}.\mathrm{862} \\ $$

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