An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be :

Question

An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be : (A) 0.01 (B) 0.001 (C) 0.002 (D) 0.003

Tardigrade Admin · Accepted Answer

AgNO3 + [Cr(H2O)4 Cl2 ]Cl arrow AgCl Excess mole = MV(ilt) = 0.01 × (100/1000) = 0.001

Q. An excess of $A g N O_{3}$ is added to $100 m L$ of a $0.01 M$ solution of dichlorotetraaquachromium(III) chloride. The number of moles of $A g Cl$ precipitated would be :

0.01

0.001

0.002

0.003

$A g N O_{3} + [C r (H_{2} O)_{4} C l_{2}] Cl \to A g Cl$
Excess mole = $M V_{(i lt)}$
$= 0.01 \times \frac{100}{1000}$
$= 0.001$

Q. An excess of AgNO3​ is added to 100mL of a 0.01M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be :