Question

What is the molarity of a solution if 300.0 ml of it contains 16.8 g of kno3? [formula weight: kno3, 101.11 amu]?

Answer 1

Answer : The molarity of a solution is 0.554 M.

Explanation : Given,

Mass of $KNO_3$ = 16.8 g

Volume of solution = 300.0 mL

Molar mass of $KNO_3$ = 101.11 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }KNO_3\times 1000}{\text{Molar mass of }KNO_3\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{16.8g\times 1000}{101.11g/mole\times 300.0mL}=0.554mole/L=0.554M$

Therefore, the molarity of a solution is 0.554 M.

Answer 2

The molarity is count by dividing the mole of the solute within 1 liter of solvent. In this case, the KNO3 is 16.8g with 101.11 g/mol molar mass. Then we need to find the mol first. The calculation would be: 16.8g / (101.11g/mol)= 0.0166 mol.

Then the molarity would be: 0.0166mol/ 0.3l= 0.0498= 0.0553 M