How many grams of $M g S O_{4}$ $MgSO_4$ are needed to prepare 125mL 0.200M magnesium sulfate?

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Answer 1 · 2017-03-01T06:25:38

$Concentration$ $"Concentration"$ $=$ $=$ $\frac{Moles of solute}{Volume of solution}$ $"Moles of solute"/"Volume of solution"$

And $mass of solute = moles \times molar mass$ $"mass of solute "=" moles"xx"molar mass"$

And so we calculate the product:

$Volume \times Concentration \times Molar mass of magnesium sulfate$ $"Volume"xx"Concentration"xx"Molar mass of magnesium sulfate"$

$=125xx10^-3cancel(L)xx0.200*cancel(mol*L^-1)xx120.37*g*cancel(mol^-1)$ , an answer with units of mass as required.

$=??*g$

How many grams of MgSO_4MgSO4MgSO_4 are needed to prepare 125mL 0.200M magnesium sulfate?