Calculate the volume in milliliters of a 1.3M zinc nitrate solution that contains 100. g of zinc nitrate ZnNO32. Be sure your answer has the correct number of significant digits.

The volume is 406 mL

Explanation:

Step 1: Data given

Molarity of a zinc nitrate solution = 1.3 M

Mass of zinc nitrate = 100 grams

Molar mass of Zn (NO3) 2 = 189.36 g/mol

Step 2: Calculate moles Zn (NO3) 2

Moles Zn (NO3) 2 = mass Zn (NO3) 2 / molar mass Zn (NO3) 2

Moles Zn (NO3) 2 = 100.0 grams / 189.36 g/mol

Moles Zn (NO3) 2 = 0.5281 moles

Step 3: Calculate volume

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.5281 moles / 1.3 M

Volume = 0.406 L = 406 mL

The volume is 406 mL