Which of the following will form negatively charged colloidal solution?

100 m L 0.1 M A g N O_{3} + 100 m L 0.1 M K I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 m L 0.2 M A g N O_{3} + 100 m L 0.1 M K I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 m L 0.1 M A g N O_{3} + 100 m L 0.2 M K I

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100 m L 0.2 M A g N O_{3} + 200 m L 0.1 M K I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $100 m L 0.1 M A g N O_{3} + 100 m L 0.2 M K I$

$Milli moles of A g N O_{3} = 100 \times 0.1 = 10$
$Milli moles of K I = 100 \times 0.2 = 20$
$K I$ is in excess.

$A g N O_{3} (a q) + K I (a q) \to A g I (s) ↓ + A g N O_{3} (a q)$
$A g I$ gets precipitated.
The precipitated silver iodide adsorbs iodide ions $(I^{-})$ from the dispersion medium and negatively charged colloidal sol is formed $A g I / I^{-}$

Therefore, correct option is (c)

Suggest Corrections

Similar questions

N a^{+}

B a^{2 +}

A l^{3 +}

[A g I]^{- 1}

Join BYJU'S Learning Program