Question

5.33 mg salt of [Cr (H20) Cl]Cl2 H2O is treated with excess AgNO3 . What is the amount of AgCl precipitated ​

Answer 1

Answer:

7.69mg of AgCl will precipitation out

Explanation:

mass of 1 mol of complex=193.5g

so number of moles=2.7*10^-5

1 mol complex gives 2 mol AgCl on treatment with excess AgNO3

so 2.7*10^-5 will give 2*2.7*10^-5

therefore the mass obtained will be number of moles into mass of 1 mole

so 2*2.7*10^-5*142.5=769.5*10^-5

which is 7.69mg