Question #120944

In a study of reaction rates, you need to dilute copper(II) sulfate solution. You take 5.00 mL of 0.050 M CuSO4(aq) and dilute this to a final volume of 100.0 mL


a) What is the final concentration of the dilute solution?



b) What mass of CuSO4(s) is present in 10.0 mL of the final dilute solution?


c) Can this final dilute solution of 10 mL be prepared directly using the pure solid? Defend your answer.

Expert's answer

Let's calculate the concentration by formula

M1V1=M2V25×0.05=M2×100M2=2.5×103M_1V_1=M_2V_2\\5\times0.05=M_2\times100\\M_2=2.5\times10^{-3}


moles of CuSO4 in 10 ml of this solution = 2.5×103×10=0.025moles2.5\times10^{-3}\times10=0.025 moles

molar mass of copper sulphate = 159.6 grams

mass of copper sulphate here = 159.6×\times 0.025=4 grams



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