The question is incomplete, here is the complete question:
Calculate the volume in milliliters of a 1.30 M zinc nitrate solution that contains 100.g of zinc nitrate [tex]Zn(NO_3)_2[/tex]. Be sure your answer has the correct number of significant digits.
Answer: The volume of solution is 406. mL
Explanation:
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Molarity of solution = 1.30 M
Given mass of zinc nitrate = 100. g
Molar mass of zinc nitrate = 189.4 g/mol
Putting values in above equation, we get:
[tex]1.3M=\frac{100\times 1000}{189.4\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100\times 1000}{189.4\times 1.3}=406.mL[/tex]
Hence, the volume of solution is 406. mL
