Question

How many moles of ions are present in an ideal solution that is produced by dissolving 22.6 g of Cu(NO 3) 2 in 323 g of water?

Answer 1

Answer: 0.360 moles

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=(22.6g)/(187.56g/mol)=0.120moles`

`Cu(NO_3)_2(aq)\rightarrow Cu^(2+)(aq)+2NO_3^-(aq)`

1 mole of
`Cu(NO_3)_2` gives = 3 moles of ions

0.120 moles of
`Cu(NO_3)_2` give=
`(3)/(1)* 0.120=0.360` moles of ions.

Thus 0.360 moles of ions are present in an ideal solution that is produced by dissolving 22.6 g of
`Cu(NO_3)_2` in 323 g of water

Answer 2

Step-by-step explanation:

Mol weight of Cu(NO 3) 2 ( trihydrate form )

= 241.6 g

22.6 g of Cu(NO 3) 2

= 22.6 / 241.6

= 1 / 10 mole ( approx )

It will ionise in water as follows

Cu(NO 3) 2 ⇄ Cu ⁺² + 2 NO₃⁻¹

Total ions per molecule = 3 , because at this dilution molecule will be fully ionised.

Total ions per 1 /10 moles = 3 / 10 moles

= .3 moles.