Question

We have an aqueous solution that contains 23% (by mass) of a hypothetical solute Z. The formula weight of the solute Z is 139 g/mol. The density of the solution is observed to be 1.3 g/mL. What is the molarity of Z in this solution

Answer 1

Answer: The molar concentration of Z in the solution is 2.15 M

Explanation:

We are given:

23% (w/w) Z in solution

This means that 23 grams of Z is present in 100 grams of solution

To calculate the volume of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.3 g/mL

Mass of solution = 100.0 g

Putting values in above equation, we get:

$1.3g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.3g/mL}=76.92mL$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of Z = 23 g

Molar mass of Z = 139 g/mol

Volume of solution = 76.92 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{23\times 1000}{139\times 76.92}\\\\\text{Molarity of solution}=2.15M$

Hence, the molar concentration of Z in the solution is 2.15 M