Question

250 ml of sodium carbonate solution contains $ 2\cdot 65 $ gm of $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} $ . If 10 ml of solution is diluted to 500 ml, the concentration of dilute acid will be?

Answer 1

Hint: Molarity of a solution can be defined as the number of moles of solute (n) present in per liter of a solution. Molarity is represented by M. Molarity of a solution is also known as the molar concentration of the solution.

Formula used: $ \text{Molarity =}\dfrac{\text{n}}{\text{V in ml}}\times 1000 $
Where, n is the number of moles,
V is the solution in milliliters.

Complete Step by step solution
Firstly, we have to find the number of moles of solute present in the solution.
We know that,
 $ \text{n}=\dfrac{\text{mass given}}{\text{Original molar mass}} $
Then, $ \text{nN}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} $ means number of moles of $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} $
Now,
 $ \text{nN}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}=\dfrac{\text{Given mass of N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}}{\text{Original molar mass of N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}} $
Molar mass of $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}=106\text{ g/mol} $
 $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}=2\left( 23 \right)+12+3\left( 16 \right) $
 $ \Rightarrow 46+12+48 $
 $ \Rightarrow 106\text{ g/mol} $
Number of moles of
 $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}=\dfrac{2\cdot 65}{106} $
  $ \Rightarrow 0\cdot 025\text{ mol} $ $ $
Thus,
 $ \text{Molarity}=\dfrac{0\cdot 025}{250}\times 1000 $
 $ \Rightarrow 0\cdot 1\text{ M} $
Now, we have to calculate the concentration of dilute acid.
Taking, $ {{\text{M}}_{1}} $ as the molarity of $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} $ and $ {{\text{M}}_{2}} $ is the molarity of dilute acid.
 $ {{\text{V}}_{1}} $ is the volume of $ \text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}} $ for dilute acid, which is 10 ml and $ {{\text{V}}_{2}} $ is the volume of dilute acid which is 500 ml.
Thus,
 $ {{\text{M}}_{1}}{{\text{V}}_{1}}={{\text{M}}_{2}}{{\text{V}}_{2}} $
Now, for the concentration of dilute acid
 $ {{\text{M}}_{2}}\text{=}\dfrac{{{\text{M}}_{1}}{{\text{V}}_{1}}}{{{\text{V}}_{2}}} $
 $ {{\text{M}}_{2}}=\dfrac{0\cdot 1\times 10}{500} $
 $ {{\text{M}}_{2}}=0\cdot 002\text{ M} $ .

Note
Molarity is a unit of concentration, $ \text{1 M =}\dfrac{1\text{ mole solute}}{1\text{ liter of solvent}} $
The molarity is $ \text{M}=\dfrac{\text{n}}{\text{V}} $ , which is also the same formula for concentration. Thus, we can say that molarity and concentration are the same terms and we can use molarity to find the concentration of solute in a solution.