How many milliliters of a 3.28 M copper (II) nitrate Cu(NO3)2 solution are needed to provide a total of 146.59 g of Cu(NO3)2?
Use 187.56 g/mol as the molar mass of solute. Solve your answer
to two decimal places.

237.80mL

Explanation:

Molarity of a solution can be calculated using the formula;

Molarity = number of moles (n) ÷ volume (V)

However, we first need to calculate the number of moles of Cu(NO3)2 by using the formula:

moles = mass/molar mass

Molar mass of Cu(NO3)2 = 187.56 g/mol

moles = 146.59 ÷ 187.56

moles = 0.782moles

Molarity = n/V

3.28 = 0.78/V

V = 0.78/3.28

V = 0.2378L

In milliliters (mL), 0.24 litres = 0.24 × 1000

= 237.80mL

elements make up everything, and they combine to make compounds.

the answer is c