Question

Define the term molarity.

Answer 1

Hint: Molarity is simply we can say that the no of moles of solute to the volume of the substance in which the solute is dissolved and is dependent on the volume of that very substance and its increase and decrease in volume has significant effect on the molarity of the solution. Now you can easily answer the statement.

Complete answer:
By the term molarity, we mean the no of moles of the solute to the total volume of the solution in liters or the volume of the solution in milliliters.
Molarity is represented by the symbol M.
The formula for the molarity in liters is as;
$molarity=\dfrac{\text{ no}\text{. of moles of the solute}}{\text{ total volume of the solution in liters}}$ --------------(1)
In milliliters;
$molarity=\dfrac{\text{ no}\text{. of moles of the solute}\times \text{1000}}{\text{ volume of the solution in milliliters}}$
Molarity of the solution is dependent on the temperature. As the temperature changes, the molarity also changes because the volume changes.
When temperature is increased, the volume of the solution increases and hence the molarity of the solution decreases because molarity and volume are inversely proportional to one another as it is clear from its formula in equation (1)
On the contrary, when temperature is decreased, the volume of the solution decreases and hence the molarity of the solution increases because molarity and volume are inversely proportional to one another as it is clear from its formula in equation (1)

Note:
Don’t get confused in the terms molarity and molality.
Molarity is defined as the no.of moles of the solute to the volume of solution in liters or milliliters whereas molality is defined as the no. of moles the solute to the total mass of the solvent in kilograms or grams.
Molarity is represented by M and molality is represented by m.
$molarity=\dfrac{\text{ no}\text{. of moles of the solute}}{\text{ total volume of the solution in liters}}$
$molality=\dfrac{\text{ no}\text{. of moles of the solute}}{\text{ Mass of the solvent in kilograms}}$