Question

A 500 mL hypertonic saline solution is labeled as consisting of 5.21 % w/w NaCl. Given that the density of salt water is 1.02 g/mL, what is the molarity of the saline solution? Molar mass of NaCl = 58.44 g/mol.

Answer 1

Answer: The molarity of saline solution is 0.909 M

Step-by-step explanation:

We are given:

5.21 w/w % NaCl

This means that 5.21 grams of NaCl is present in 100 grams of solution

To calculate mass of a substance, we use the equation:

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

Density of solution = 1.02 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

`1.02g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=(100g)/(1.02g/mL)=98.04mL`

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

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

Given mass of NaCl = 5.21 g

Molar mass of NaCl = 58.44 g/mol

Volume of solution = 98.04 mL

Putting values in above equation, we get:

`\text{Molarity of solution}=(5.21* 1000)/(58.44g/mol* 98.04)\\\\\text{Molarity of solution}=0.909M`

Hence, the molarity of saline solution is 0.909 M