Calculate each of the following quantities:

(a) Mass in kilograms of 4.6×1021 molecules of NO2

(b) Moles of Cl atoms in 0.0615 g of C2H4C12

(c) Number of H− ions in 5.82 g of SrH2

Solution 12P:

Here, we are going to calculate the following quantities.

Step 1:

(a) Mass in kilograms of 4.6×1021 molecules of NO2

Here, we have to calculate the mass in kg of 4.6×1021 molecules of NO2

We know that,

The molar mass of NO2 = 46.01g/mol

Given that,

Number of the solute = 4.6×1021 molecules

Therefore,

1.0 mole of NO2 = 6.022 x1023 molecules of NO2

Thus,

The mole of NO2 =4.6×1021 molecules = 7.638 10-3 moles of NO2

We know that,

1.0 mole of NO2 = 46.01 g

Therefore,

The mass of NO2 = 7.638 10-3 moles of NO2 = 351.4210-3 g

We know,

1000 g =1.0 kg

Therefore, the mass of NO2 in kg = 351.4210-3 g = 3.5110-4 kg

Step 2:

(b) Moles of Cl atoms in 0.0615 g of C2H4C12

Here, we have to calculate the moles of Cl atoms in 0.0615 g of C2H4C12

We know that,

The molar mass of C2H4C12 =98.95 g/mol

Given that,

Mass of the solute = 0.0615 g

Thus,

98.95 g of C2H4C12 = 1.0 mole

Therefore,

The moles of C2H4C12 =0.0615 g = 6.215 x 10–4 moles of C2H4C12

The dissociation of C2H4C12 gives:

C2H4C122C + 4H + 2 Cl

Therefore,

1.0 mole of C2H4C12= 2.0 moles of Cl atoms

Thus,

The moles of Cl atoms = 6.215 x 10–4 moles of C2H4C12

= 6.215 x 10–4 moles of Cl atoms

Therefore, 6.215 x 10–4 moles of Cl atoms is in 0.0615 g of C2H4C12.