Question

calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water.

Deg. of dissociation, alpha = 0.065

Kf = 1.86 K Kg/mol



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Answer 1

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depression in freezing point, ∆Tf = 0.65 K

step-by-step explanation :

STEP 1. CALCULATING THE MOLALITY

mass of the solute = 10 g

Molar mass of the solute( CH3CH2CHClCOOH )

= (12×4)+(1×7)+(35.5)+(16×2)

= 48 +7+ 35.5 +32

= 122.5 g/mol

Now,

MOLALITY

= (mass of solute/molar mass of solute)/Mass of solvent in Kg

= 10 g/(122.5 g/mol)(0.25 Kg)

m = 0.326 mol / Kg

STEP 2. CALCULATING VAN'T HOFF FACTOR (i)

Note :-

The reaction is given in attachment.

seeing in the reaction,

total no. of moles after dissociation

= (1- α) + 2•α

= 1+α

Van't Hoff Factor,

(i) = total no. of molesafter dissociation/no. of moles before dissociation

=> (i) = (1+α)/1

=> (i) = 1+α

=> (i) = 1+ 0.065

=> (i) = 1.065

STEP 3. CALCULATING DEPRESSION IN FREEZING POINT

∆ Tf = i • Kf • m

= (1.065) × (1.86 K kg/mol) × (0.326mol/kg)

=> ∆ Tf = 0.65 K

Answer 2

$\textbf{Answer is in Attachment !!}$