Question

Calculate the molarity of the following solution:
a.4g of caustic soda is dissolved in 200mL of the solution.
b.5.3g of anhydrous sodium carbonate dissolved in 100mL of solution.
c.0.365g of pure HCl gas dissolved in 50mL of solution.

Answer 1

We can calculate molarity of the solution from the volume of the solution and moles of compounds/substances. We can calculate the moles of the substances/compounds using their molar mass.

Formula used: We can define Molarity as the mass of solute in one liter of solution. Molarity is the desired concentration unit for stoichiometry calculations. The formula is,
${\text{Molarity}} = \dfrac{{{\text{Moles of solute}}\left( {{\text{in moles}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$

Complete step by step answer:
Given data contains
Mass of caustic soda is 4g.
Volume of the solution is 200mL
The molar mass of caustic soda is 40g/mol.
From the mass and molar mass of caustic soda, we can calculate the moles of caustic soda.
${\text{Moles}} = \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
Let us substitute the values of mass and molar mass of caustic soda to get the moles of caustic soda.
$Moles = \dfrac{{4g}}{{40g/mol}}$
On simplifying we get,
$Moles = 0.1mol$
The moles of caustic soda are 0.1moles.
From the moles of caustic soda and the volume of the solution we can calculate molarity of the solution.
The volume of solution in litres is 0.2L.
We know the formula to calculate molarity is,
${\text{Molarity}} = \dfrac{{{\text{Moles of solute}}\left( {{\text{in moles}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
We have to substitute the moles of caustic soda and volume to get molarity of the solution.
${\text{Molarity}} = \dfrac{{{\text{Moles of caustic soda}}}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
Substituting the known values we get,
$Molarity = \dfrac{{0.1mol}}{{0.2L}}$
On simplifying we get,
$Molarity = 0.5M$
The molarity of the solution is $0.5M$.
Given data contains
Mass of anhydrous sodium carbonate is 5.3g.
Volume of the solution is 100mL
The molar mass of anhydrous sodium carbonate is 106g/mol.
From the mass and molar mass of anhydrous sodium carbonate, we can calculate anhydrous sodium carbonate.
${\text{Moles}} = \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
Let us substitute the values of mass and molar mass of anhydrous sodium carbonate to get the moles of anhydrous sodium carbonate.
$Moles = \dfrac{{5.3g}}{{106g/mol}}$
On simplifying we get,
$Moles = 0.05mol$
The moles of anhydrous sodium carbonate are 0.05moles.
From the moles of anhydrous sodium carbonate and the volume of the solution we can calculate molarity of the solution.
The volume of solution in litres is 0.1L.
We know the formula to calculate molarity is,
${\text{Molarity}} = \dfrac{{{\text{Moles of solute}}\left( {{\text{in moles}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
We have to substitute the moles of anhydrous sodium carbonate and volume to get molarity of the solution.
${\text{Molarity}} = \dfrac{{{\text{Moles of anhydrous sodium carbonate}}}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
$Molarity = \dfrac{{0.05mol}}{{0.1L}}$
On simplifying we get,
$Molarity = 0.5M$
The molarity of the solution is $0.5M$.
Given data contains
Mass of pure HCl gas is 0.365g.
Volume of the solution is 50mL
The molar mass of pure HCl gas is 36.5g/mol.
From the mass and molar mass of pure HCl gas, we can calculate the moles of pure HCl gas.
${\text{Moles}} = \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$
Let us substitute the values of mass and molar mass of pure HCl gas to get the moles of pure HCl gas.
$Moles = \dfrac{{0.365g}}{{36.5g/mol}}$
On simplifying we get,
$Moles = 0.01mol$
The moles of pure HCl gas are 0.1moles.
From the moles of pure HCl gas and the volume of the solution we can calculate molarity of the solution.
The volume of solution in litres is 0.05L.
We know the formula to calculate molarity is,
${\text{Molarity}} = \dfrac{{{\text{Moles of solute}}\left( {{\text{in moles}}} \right)}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
We have to substitute the moles of pure HCl gas and volume to get molarity of the solution.
${\text{Molarity}} = \dfrac{{{\text{Moles pure HCl gas}}}}{{{\text{Volume of solution}}\left( {{\text{in litres}}} \right)}}$
$Molarity = \dfrac{{0.01mol}}{{0.05L}}$
On simplifying we get,
$Molarity = 0.2M$
The molarity of the solution is $0.2M$.

Note:
From molarity, we can also calculate the mass percentage, molality. An example of calculation of mass percentage from molarity is shown below.
Example:
The concentration of the phosphoric acid expressed in mass percentage has to be calculated.
Given,
Molarity of the solution is $0.631M$
Density of the solution is $1.031g/ml$
The grams of the phosphoric acid are calculated using the molar mass.
Grams of phosphoric acid=$0.631mol \times \dfrac{{97.994g}}{{1mol}} = 61.8342g$
The mass of the solution is calculated from the density of solution
Mass of the solution=$1L \times \dfrac{{1.031g}}{{1ml}} \times \dfrac{{1000ml}}{{1L}} = 1031g$
The concentration of the solution is,
Mass percentage$= \dfrac{{{\text{Grams of solute}}}}{{{\text{Grams of solution}}}} \times 100\%$
Concentration of the solution=$\dfrac{{61.8342g}}{{1031g}} \times 100\%$
Concentration of the solution=$5.9974\%$
The concentration of the solution in mass percentage is $5.9974\%$.