Question

How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

Answer 1

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = $\frac{no. of moles}{Volume of solution in liter}$

            0.0800 M = $\frac{no. of moles}{0.05 L}$

            no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = $\frac{mass in grams}{molar mass}$

             mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

                                       = $1.6 mol \times 249.68 g/mol$

                                       = 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.