Question

Class 12th boards question


Calculate the depression in freezing point of water when 20.0 g of CH3CH2CHClCOOH is added to 500 g of water.

given,

Ka = 1.4 × 10^-3

Kf = 1.86 K kg/mol

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Answer 1

$\huge\bf\mathscr\pink{Your\: Answer}$

≈ 0.65 K

step-by-step explanation:

Given,

mass of CH3CH2CHClCOOH = 20.0 g

mass of water = 500 g

Ka = 1.4 × ${10}^{-3}$ K•kg /mol

Kb = 1.86 K•kg/mol

now,

molar mass of CH3CH2CHClCOOH

= 4×12 + 7×1 + 35.5 + 2×16

= 122.5 g/mol

So,

No. of moles of CH3CH2CHClCOOH

= 20/122.5

= 0.1632 mol

Now,

molality of solution

= mass if solute × 100/mass of solvent (g)

=> m = 0.1632× 1000/500

=> m = 0.3264 m

Now,

in the raction in the attachment,

total no. of moles at equilibrium

= C - Cα + Cα - Cα

= C - Cα

= C (1-α)

where,

α is the degree of the dissociation of the acid

Now,

Ka = ${(Cα)}^{2}$ / C(1-α)

=> Ka = C ${α}^{2}$

(°.° it is a weak electrolyte, 1-α = 1 )

=> α = √(Ka/C)

=> α = √(1.4× ${10}^{-3}$ /0.3264 )

=> α = 0.065

Now,

we know that,

α = i -1/n -1

But,

n = 2 ,for the given acid

(because 1 mol gives 2 particles on dissociation)

.°. 0.065 = i-1/2-1

=> i = 1+ 0.065 = 1.065

Now,

∆Tf = i•Kf•m

= 1.065 × 1.86 K kg/mol × 0.3264 m

=> ∆Tf = 0.6465 ≈ 0.65 K

Answer 2

Molar Mass of $CH_{3}CH_{2}CHClCOOH$

= 12 + 3 + 12 + 2 + 12 + 1 + 35.5 + 12 + 16 + 16 + 1

= 122.5 g/mol

Number of moles of $CH_{3}CH_{2}CHClCOOH$ in 10 gms.

$\dfrac{10}{122.5}$ = $\dfrac{8.16}{100}$

= 0.0816 mol


» $\underline{\underline{Molarity}}$

$\dfrac{Number\:of\:moles\:of\:solute}{Volume\:of\: solution\:in\:litre}$

= $\dfrac{0.0816\:\times\:1000}{250}$

= $\textbf{0.3265 mol/kg}$


» $\underline{\underline{Degree\:of\: Dessociation}}$

$CH_{3}CH_{2}CHClCOOH$ <---> $CH_{3}CH_{2}CHClCO{O}^{-}$ + ${H}^{+}$

• [ In Attachment ]

$k_{a}$ = $\dfrac{C \alpha \: \times \: V \alpha }{C(1 \: - \: \alpha )}$

= $\dfrac{C { \alpha }^{2} }{V(1 \: - \: \alpha )}$

a = $\sqrt{\dfrac{ka}{C}}$

= $\dfrac{1.4\:\times\:{10}^{-3}}{0.3265}$

= $\textbf{0.065}$


» $\underline{\underline{Van't\:Hoff\:Factor}}$

$\dfrac{1 \: + \: \alpha }{1}$

= 1 + 0.065

= 1.065

∆$T_{f}$ = i $k_{f}$ mv

= 1.065 × 1.86 × 0.3265

= 0.647

= $\underline{\underline{\bold{0.65 K}}}$ (Approx.)