Calculate each of the following quantities:

(a) Total number of ions in 38.1 g of SrF2

(b) Mass in kilograms of 3.58 mol of CuCl2·2H2O

(c) Mass in milligrams of 2.88×1022 formula units of Bi(NO3)3·5H2O

Solution 14P:

Step 1:

(a) Total number of ions in 38.1 g of SrF2

Here, we have to calculate the total number of ions in 38.1 g of SrF2.

The molar mass of SrF2 = 125.62 g/mol

Thus,

The amount of SrF2 in moles = 38.1 g of SrF2 = 0.30329 moles of SrF2

Given that, total moles of ions in SrF2 = 1.0 mol of Sr2++ 2.0 moles of F- =3.0 moles

Therefore,

Total moles of ions in 0.30329 moles of SrF2

= (0.30329 moles of SrF2) = 0.90988 mol ions

We know that,

1.0 mole of ions = 6.022ions

Thus, the number of total ions = ( 0.90988 mol ions) = 5.479 ions

Therefore, 5.479 ions are in 38.1 g of SrF2

Step 2:

(b) Mass in kilograms of 3.58 mol of CuCl2·2H2O

Here, we have to calculate the total mass in kilograms of 3.58 mol of CuCl2·2H2O

The molar mass of CuCl2·2H2O = 170.48 g/mol

1000.0 g =1.0 kg

Given that,

Mass of CuCl2·2H2O =3.58 g

Therefore,

The mass of CuCl2·2H2O = 3.58 g CuCl2·2H2O

= 0.610 kg

Thus, the mass in kilograms of 3.58 mol of CuCl2·2H2O is 0.610 kg.