A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of BaCl 2, AlCl 3 and NaCl required to coagulate the given colloid is in the order

Question

A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of BaCl 2, AlCl 3 and NaCl required to coagulate the given colloid is in the order (A) NaCl > BaCl 2> AlCl 3 (B) BaCl 2> AlCl 3> NaCl (C) AlCl 3= NaCl = BaCl 2 (D) AlCl 3> BaCl 2> NaCl

Tardigrade Admin · Accepted Answer

Hardy Schulze Law: The effective ions of the electrolytes in bringing about coagulation are those which carry a charge opposite to that of colloidal particles. Greater the valency of coagulating or fluctuating ion, greater is its power to bring about coagulation.

When colloidal solution is subjected to electrical field, coagulating power should be in the order of

A lC l_{3} > B a C l_{2} > N a Cl

as thier valency is increasing.

Q. A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of $B a C l_{2}, A lC l_{3}$ and $N a Cl$ required to coagulate the given colloid is in the order

$N a Cl > B a C l_{2} > A lC l_{3}$

$B a C l_{2} > A lC l_{3} > N a Cl$

$A lC l_{3} = N a Cl = B a C l_{2}$

$A lC l_{3} > B a C l_{2} > N a Cl$

Q. A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of BaCl2​,AlCl3​ and NaCl required to coagulate the given colloid is in the order

NaCl>BaCl2​>AlCl3​

BaCl2​>AlCl3​>NaCl

AlCl3​=NaCl=BaCl2​

AlCl3​>BaCl2​>NaCl

Q. A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of $B a C l_{2}, A lC l_{3}$ and $N a Cl$ required to coagulate the given colloid is in the order

$N a Cl > B a C l_{2} > A lC l_{3}$

$B a C l_{2} > A lC l_{3} > N a Cl$

$A lC l_{3} = N a Cl = B a C l_{2}$

$A lC l_{3} > B a C l_{2} > N a Cl$