Calculate the volume in milliliters of a 2.52 mol/L zine nitrate solution that contains 50.0 g of zine nitrate (Zn(NO3)2) Be sure your answer has the correct number of significant digits.

Answer 1

To calculate the volume of the solution, we need to use the formula:

Volume (mL) = (mass of solute / molar mass of solute) / concentration

First, we need to calculate the molar mass of zinc nitrate:

Zn(NO3)2
Zinc (Zn) = 65.38 g/mol
Nitrogen (N) = 14.01 g/mol
Oxygen (O) = 16.00 g/mol

Molar mass of Zn(NO3)2 = 65.38 + (14.01 + 3(16.00)) = 65.38 + 62.03 = 127.41 g/mol

Next, we substitute the values into the formula:

Volume (mL) = (50.0 g / 127.41 g/mol) / 2.52 mol/L
Volume (mL) = (0.3926 mol) / 2.52 mol/L
Volume (mL) = 0.1558 L
Volume (mL) = 155.8 mL

Therefore, the volume of the 2.52 mol/L zinc nitrate solution that contains 50.0 g of zinc nitrate is 155.8 mL.