Question

Calcium nitrate dissolves as follows: . how many moles of ca(no 3) 2 are required to make 2.0 liters of solution in which the no 3 - concentration is 0.20 m?

Answer 1

Answer : The number of moles of
`Ca(NO_3)_2` required are, 0.2 moles

Explanation :

First we have to calculate the moles of
`NO_3^-`.

`\text{Moles of }NO_3^-=\text{Molarity of }NO_3^-* \text{Volume of solution}=0.20M* 2.0L=0.4mole`

Now we have to calculate the moles of
`Ca(NO_3)_2`.

The dissociation chemical reaction will be,

`Ca(NO_3)_2\rightarrow Ca^(2+)+2NO_3^-`

From this reaction we conclude that,

As, 2 moles of
`NO_3^-` obtained from 1 mole of
`Ca(NO_3)_2`

So, 0.4 moles of
`NO_3^-` obtained from
`(0.4)/(2)=0.2` mole of
`Ca(NO_3)_2`

Therefore, the number of moles of
`Ca(NO_3)_2` required are, 0.2 moles

Answer 2

If the volume of the solution is 2.0 liters and the concentration of NO3 - is 0.20 m, the number of mol of NO3- in the solution would be:
mol= volume * concentration= 2 liters * 0.2 mol/liter= 0.4 mol

The equation of ion formed by ca(no 3) 2 would be:
Ca(NO3)2 ==> 1 Ca + 2 NO3
For every 1 mol of Ca(NO3)2 will produce 2 mol of NO3-. So, the amount of Ca(NO3)2 needed to make 0.4 mol of NO3- would be: 0.4 mol / (2/1)= 0.2 mol