Question

alculate the volume in milliliters of a M zinc nitrate solution that contains of zinc nitrate . Round your answer to significant digits

Answer 1

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
`Zn(NO_3)_2`. Be sure your answer has the correct number of significant digits.

Answer: The volume of solution is 406. mL

Step-by-step explanation:

To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

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:

`1.3M=\frac{100* 1000}{189.4* \text{Volume of solution}}\\\\\text{Volume of solution}=(100* 1000)/(189.4* 1.3)=406.mL`

Hence, the volume of solution is 406. mL