Question

The quantity of Ceric ammonium sulphate
required to prepare 0.1 N solution of Ceric
ammonium sulphate 1 litre (1 marks)​

Answer 1

Answer:

sorry I didn't understand but I will try

Answer 2

Answer:

To make 0.1 N solution of $[(NH_4)_2Ce(SO_4)_4]$ is 60g.

Explanation:

Ceric Ammonium Sulphate is $[(NH_4)_2Ce(SO_4)_4]$ .

Normality = No of gram equivalents of the compound/ Amount of solvent(in litres)

$N= \frac{x}{1} \\0.1 = x\\ 0.1 moles =x$

No of gram equivalents of ceric ammonium sulphate needed for 0.1N solution is 0.1 moles.

No of equivalents = Weight of the compound/equivalent weight

Equivalent weight = molecular weight × n- factor of the compound

n-factor is the valency factor.

n- factor for $[(NH_4)_2Ce(SO_4)_4]$ is 1

so Equivalent weight = Molecular weight

Molecular weight of $[(NH_4)_2Ce(SO_4)_4]$ is 600gms.

No of gram equivalents = Weight of the compound/equivalent weight

Weight of the compound = x

$0.1 = \frac{x}{600} \\60gms =x$

60 grams of $[(NH_4)_2Ce(SO_4)_4]$ is required to make a 0.1N solution.