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Molecular weight of NaCl determined by freezing point depression of its 0.5 %aqueous solution is 30. The Apparent degree of dissociation of NaCl is..... Ans is 0.95... Plz explain the solution

Molecular weight of NaCl determined by freezing point depression of its 0.5 %aqueous solution is 30. The Apparent degree of dissociation of NaCl is..... 
Ans is 0.95... Plz explain the solution 

Grade:12

1 Answers

Umakant biswal
5349 Points
7 years ago
@ nik 
can u able to recall a formula in the solution chapter , the formula is 
delta Tf = i*Kf*m.
where t(f) is the freezing point , i is the vant hoff factor kf is the depression constant and m is the molecular mass, 
this formula  will give iinformation about dissociation.
and i = normal molar mass/observed molar mass = 58.5/30 = 1.95 = 1+ alpha
so, by applying the value and calculating we have 
alpha = 0.95 
HOPE IT CLEARS YOUR DOUBT 

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