What is the concentration (molarity) of a solution created by dissolving 81.1 grams of Mg3P2 to create a 2.5 solution?

Chemistry

1 answer:

Vaselesa [24]1 year ago

8 0

Answer:

The concentration of Mg3P2 solution = 0.24mol/L

Explanation:

Mass = 81.1g

Volume = 2.5L

Concentration (C) of a solution = $\frac{Moles (n)}{Volume (V)}$

but moles = $\frac{mass}{Molar mass}$ and Molar mass of Mg3P2 = 134.88g/mol

Moles = $\frac{81.1g}{134.88g/mol}$

= 0.60mol

hence the concentration can be calculated by;

C = $\frac{0.60mol}{2.5L} \\$

= 0.24mol/L or 0.24M

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