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

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:

$1.02g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.02g/mL}=98.04mL$

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

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

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:

$\text{Molarity of solution}=\frac{5.21\times 1000}{58.44g/mol\times 98.04}\\\\\text{Molarity of solution}=0.909M$

Hence, the molarity of saline solution is 0.909 M