The van't Hoff factor for a very dilute solution of Fe2(SO4)3 is

Question

The van't Hoff factor for a very dilute solution of Fe2(SO4)3 is (A) 9 (B) 5 (C) 24 (D) none of these

Tardigrade Admin · Accepted Answer

F e2( SO 4)3 is very dilute solution ionises as Fe 2( SO 4)3 leftharpoons 2 Fe 3++3 SO 42-. ∴ one formula unit of ferric sulphate gives 5 ions in solution. Hence, i=5.

Q. The van't Hoff factor for a very dilute solution of $F e_{2} (S O_{4})_{3}$ is

9

5

24

none of these

$F e_{2} (S O_{4})_{3}$ is very dilute solution ionises as
$F e_{2} (S O_{4})_{3} ⇌ 2 F e^{3 +} + 3 S O_{4}^{2 -}$ .
$∴$ one formula unit of ferric sulphate gives 5 ions in solution. Hence, $i = 5$ .

Q. The van't Hoff factor for a very dilute solution of Fe2​(SO4​)3​ is

9

5

24

none of these

Fe2​(SO4​)3​ is very dilute solution ionises as Fe2​(SO4​)3​⇌2Fe3++3SO42−​. ∴ one formula unit of ferric sulphate gives 5 ions in solution. Hence, i=5.

Q. The van't Hoff factor for a very dilute solution of $F e_{2} (S O_{4})_{3}$ is

$F e_{2} (S O_{4})_{3}$ is very dilute solution ionises as
$F e_{2} (S O_{4})_{3} ⇌ 2 F e^{3 +} + 3 S O_{4}^{2 -}$ .
$∴$ one formula unit of ferric sulphate gives 5 ions in solution. Hence, $i = 5$ .