Question

If a solution contains 4.00 moles of MgCl2 in 40.0 moles of water, what is the mass percent concentration of MgCl2? Answer to three significant figures.

Answer 1

Given:-

→ Number of moles of MgCl₂ = 4.00

→ Number of moles of water = 40.0

To find:-

→ Mass percent concentration of MgCl₂.

Solution:-

[Mg = 24 g/mol ; Cl = 35.5 g/mol]

Molar mass of MgCl₂ :-

= 24 + 35.5×2

= 24 + 71

= 95 g/mol

[H = 1 g/mol ; O = 16 g/mol]

Molar mass of water (H₂O) :-

= 1×2 + 16

= 18 g/mol

________________________________

Mass of MgCl₂ in the solution :-

= No of moles × Molar mass

= 4 × 95

= 380g

Mass of water in the solution :-

= No of moles × Molar mass

= 40 × 18

= 720g

________________________________

Mass percent concentration of MgCl₂ :-

= 380/[380 + 720] × 100

= 380/1100 × 100

= 38000/1100

= 34.5%

Thus, required concentration is 34.5% .

Answer 2

Given:-

→ Number of moles of MgCl₂ = 4.00

→ Number of moles of water = 40.0

To find:-

→ Mass percent concentration of MgCl₂.

Solution:-

[Mg = 24 g/mol ; Cl = 35.5 g/mol]

Molar mass of MgCl₂ :-

= 24 + 35.5×2

= 24 + 71

= 95 g/mol

[H = 1 g/mol ; O = 16 g/mol]

Molar mass of water (H₂O) :-

= 1×2 + 16

= 18 g/mol

________________________________

Mass of MgCl₂ in the solution :-

= No of moles × Molar mass

= 4 × 95

= 380g

Mass of water in the solution :-

= No of moles × Molar mass

= 40 × 18

= 720g

________________________________

Mass percent concentration of MgCl₂ :-

= 380/[380 + 720] × 100

= 380/1100 × 100

= 38000/1100

= 34.5%

$Thus \: required \ \\ \\ : \: concentration \: \\ \\ 34.5\%$