0.01 M solution has 3 % sucrose, 4 % BaCl2, 2 % KCl and 1 % FeCl3. Then the correct order of freezing point at 1 atm pressure

Question

0.01 M solution has 3 % sucrose, 4 % BaCl2, 2 % KCl and 1 % FeCl3. Then the correct order of freezing point at 1 atm pressure (A) Sucrose < KCl< BaCl2 < FeCl3 (B) FeCl3 < BaCl2 < KCl< Sucrose (C) BaCl2 < FeCl3 < Sucrose < KCl (D) KCl< Sucrose < FeCl3 < BaCl2

Tardigrade Admin · Accepted Answer

Depression in freezing point is directly proportional to van't Hoff factor i and i is directly proportional to number of ions furnished by a molecule. Higher the ions furnished, higher is the depression in freezing point and so lower is the freezing point
Compound i
Sucrose = 1

B a C l_{2} = 3

K Cl = 2

F e C l_{3} = 4

So, the order of freezing point is

F e C l_{3} < K Cl < B a C l_{2} < sucrose

Q. $0.01 M$ solution has $3%$ sucrose, $4% B a C l_{2}$ , $2% K Cl$ and $1% F e C l_{3}$ . Then the correct order of freezing point at $1$ atm pressure

$S u crose < K Cl < B a C l_{2} < F e C l_{3}$

$F e C l_{3} < B a C l_{2} < K Cl < S u crose$

$B a C l_{2} < F e C l_{3} < S u crose < K Cl$

$K Cl < S u crose < F e C l_{3} < B a C l_{2}$

Q. 0.01M solution has 3% sucrose, 4%BaCl2​, 2%KCl and 1%FeCl3​. Then the correct order of freezing point at 1 atm pressure

Sucrose<KCl<BaCl2​<FeCl3​

FeCl3​<BaCl2​<KCl<Sucrose

BaCl2​<FeCl3​<Sucrose<KCl

KCl<Sucrose<FeCl3​<BaCl2​

Q. $0.01 M$ solution has $3%$ sucrose, $4% B a C l_{2}$ , $2% K Cl$ and $1% F e C l_{3}$ . Then the correct order of freezing point at $1$ atm pressure

$S u crose < K Cl < B a C l_{2} < F e C l_{3}$

$F e C l_{3} < B a C l_{2} < K Cl < S u crose$

$B a C l_{2} < F e C l_{3} < S u crose < K Cl$

$K Cl < S u crose < F e C l_{3} < B a C l_{2}$