Answer:
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
