The van’t Hoff factor i for an infinitely dilute solution of NaHSO4 is:

Question

The van’t Hoff factor i for an infinitely dilute solution of NaHSO4 is: (A) 1/2 (B) 1/3 (C) 3 (D) 2

Tardigrade Admin · Accepted Answer

If the solution is infinitely dilute, NaHSO4 would dissociate completely ∴ NaHSO4 → Na++HSO-4 HSO-4 would further dissociate as: HSO-4 → H+ +SO-4 i = 3

Q. The van’t Hoff factor $i$ for an infinitely dilute solution of $N a H S O_{4}$ is:

1/2

1/3

3

2

If the solution is infinitely dilute, $N a H S O_{4}$ would dissociate completely
$∴ N a H S O_{4} \to N a^{+} + H S O_{4}^{-}$
$H S O_{4}^{-}$ would further dissociate as:
$H S O_{4}^{-} \to H^{+} + S O_{4}^{-}$
$i = 3$

Q. The van’t Hoff factor i for an infinitely dilute solution of NaHSO4​ is:

1/2

1/3

3

2

If the solution is infinitely dilute, NaHSO4​ would dissociate completely ∴NaHSO4​→Na++HSO4−​ HSO4−​ would further dissociate as: HSO4−​→H++SO4−​ i=3

Q. The van’t Hoff factor $i$ for an infinitely dilute solution of $N a H S O_{4}$ is:

If the solution is infinitely dilute, $N a H S O_{4}$ would dissociate completely
$∴ N a H S O_{4} \to N a^{+} + H S O_{4}^{-}$
$H S O_{4}^{-}$ would further dissociate as:
$H S O_{4}^{-} \to H^{+} + S O_{4}^{-}$
$i = 3$