Answer:
0.105 mol/dm³
Explanation:
From the question given above, the following data were obtained:
Mass of MgCl₂ = 10 g
Volume of solution = 1 dm³
Concentration of MgCl₂ =?
Next, we shall determine the number of mole present in 10 g of MgCl₂. This can be obtained as follow:
Mass of MgCl₂ = 10 g
Molar mass of MgCl₂ = 24 + (2 ×35.5)
= 24 + 71
= 95 g/mol
Mole of MgCl₂ =?
Mole = mass /Molar mass
Mole of MgCl₂ = 10 / 95
Mole of MgCl₂ = 0.105 mole
Finally, we shall determine the concentration of MgCl₂. This can be obtained as follow:
Mole of MgCl₂ = 0.105 mole
Volume of solution = 1 dm³
Concentration of MgCl₂ =?
Concentration = mole / Volume
Concentration of MgCl₂ = 0.105 / 1
Concentration of MgCl₂ = 0.105 mol/dm³
Concentration is [tex]0.105\ mol/dm^{3}[/tex]
Given:-
Mass of MgCl₂ = 10 g
Volume of solution = [tex]1\ dm^{3}[/tex]
Concentration of MgCl₂ =?
[tex]Molar mass of MgCl_2 = 24 + (2\times35.5)\\= 24 + 71\\= 95 g/mol[/tex]
Mole of MgCl₂ =?
The number of moles is calculated by the equation as follows:-
[tex]Mole = mass /Molar massMole of MgCl_2= 10 / 95Mole of MgCl_2 = 0.105 mole[/tex]
Determine the concentration of [tex]MgCl_2[/tex] as follows:-
Mole of MgCl₂ = 0.105 mole
Volume of solution = 1 dm³
Concentration of MgCl₂ =?
[tex]Concentration = mole / Volume\\Concentration\ of\ MgCl_2 = 0.105 / 1\\Concentration \ of\ MgCl_2= 0.105 mol/dm^{3}[/tex]
Hence the concentration of 10g of magnesium chloride is [tex]0.105 mol/dm^{3}[/tex].
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