Question

A chemist has 4.0 g of silver nitrate and needs to prepare 2.0 L of a 0.010 M solution. Will there be enough silver nitrate? If so, how much silver nitrate will be left over?

Answer 1

molarity of a solution is defined as the number of moles of solute dissolved in 1 L solvent.

The molarity required in the solution to be prepared - 0.010 M

the volume of the solution to be prepared  - 2.0 L

In 1 L of solution there should be 0.010 mol

Therefore in 2.0 L solution, number of moles required - 0.010 mol/L x 2.0 L

the number of silver nitrate moles required - 0.020 mol

Molar mass of silver nitrate - AgNO₃ - 169.9 g/mol

Mass of silver nitrate needed to make solution - 0.020 mol x 169.9 g/mol  

Mass required - 3.4 g

chemist had 4.0 g and 3.4 g needed to make the solution

leftover solid AgNO₃ after the reaction is 4.0 - 3.4 = 0.6 g of AgNO₃

Answer 2

Answer:

4gm of Ag NO3

Molecular mass of Ag NO3 is 170gm

no.of moles =4/170=0.024

2L of 0.01M ,

$M = \frac{n}{v}$

n=M×v

=0.01×2=0.02 moles

The excess moles=0.004

Mass of Ag NO3 left=0.004×170

=0.68gm