Answer:
The concentration of Mg3P2 solution = 0.24mol/L
Explanation:
Mass = 81.1g
Volume = 2.5L
Concentration (C) of a solution = [tex]\frac{Moles (n)}{Volume (V)}[/tex]
but moles = [tex]\frac{mass}{Molar mass}[/tex] and Molar mass of Mg3P2 = 134.88g/mol
Moles = [tex]\frac{81.1g}{134.88g/mol}[/tex]
= 0.60mol
hence the concentration can be calculated by;
C = [tex]\frac{0.60mol}{2.5L} \\[/tex]
= 0.24mol/L or 0.24M
