Question

A laboratory procedure calls for making 600.0mL of a 1.1M $KN{{O}_{3}}$ solution. How much $KN{{O}_{3}}$in grams is needed?

Answer 1

Hint: The answer lies in the concept of basic calculation that includes the calculation of molar weights of each atom in $KN{{O}_{3}}$ and then adding it together followed by calculation of weight of $KN{{O}_{3}}$ in grams.

Complete answer:
We have studied the very common topic which explains about the calculation of several physical quantities like molarity, molality, normality and related quantities and also their molecular weight calculation.
Let us now see these concepts in detail and approach to the required answer.
Now, as per the data given we shall calculate the molecular weight of the compound that is $KN{{O}_{3}}$ and this can be done as follows,
- Molar mass of each atom in the given compound is to be calculated stepwise.
- Firstly let us see the molar weight of potassium followed by nitrogen and oxygen and they are
\[{{M}_{K}}=39.098g/mol\]
\[{{M}_{N}}=14.007g/mol\]
\[{{M}_{O}}=16g/mol\]
Now, we shall add all these values by multiplying with their respective number of atoms they are present as in the given compound and that will be,
\[{{M}_{KN{{O}_{3}}}}=39.098+14.007+3\times 16\approx 101.102g/mol\]
The mass of oxygen is multiplied by 3 because of the presence of 3 oxygen atoms in the compound $KN{{O}_{3}}$
Now, the total grams of $KN{{O}_{3}}$ present in 1.1M solution will be,
\[1.1mol{{L}^{-1}}\times \dfrac{1}{1000}\times 600\times 101.102g/mol=66.73g\]
Hence, a total of $KN{{O}_{3}}$66.73g of is required making 600mL of solution.

Note:
Note that molarity is also known as the molar concentration of a solution and this quantity is the most common unit used to measure the concentration of a solution and this quantity can be used for the calculation of volume of solvent or the amount of solute.