Question

Make 50.0 mL of a solution that is both 0.15 M in malonic acid, CH2(CO2H)2 and 0.020 M in MnSO4. Calculate how many grams of malonic acid to make 50.0 ML of 0.15 M malonic acid an dhow many grams of MnSO4 + H2O needed to make 50.0 mL of 0.020 M manganese(II) sulfate solution. Note for a hydrate, the water molecule in the formula is part of what is weighed out and must be included in the molar mass calculation.

Please show all work.

Answer 1

Molarity = Moles / Liter

Moles = Molarity x Liter

Now,

50.0 mL of 0.15 M malonic acid;

Moles of malonic acid = 0.15 M x (50/1000) L = 0.0075 moles

Molar mass of malonic acid = 104 g/mol

So, mass of 1 mole malonic acid = 104 g

mass of 0.0075 moles malonic acid = 0.0075 x 104 g = 0.78 g

and

50.0 mL of 0.020 M MnSO4.H2O

Moles of MnSO4.H2O = 0.020 M x (50/1000) L = 0.001 moles

Molar mass of MnSO4.H2O = 169 g/mol

So, mass of 1 mole MnSO4.H2O = 169 g

mass of 0.001 moles MnSO4.H2O = 0.001 x 169 g = 0.169 g