Question

If $ 100ml $ of each of $ 1M $ $ AgN{O_3} $ and $ 1M $ $ NaCl $ are mixed. The nitrate ion concentration in the resulting solution is:
A. $ 1M $
B. $ 0.25M $
C. $ 0.75M $
D. $ 0.5M $

Answer 1

Hint: Concentration of solution is defined in various methods such as molarity, molality, mole fraction and so on. Here we have to calculate the molarity of nitrate ions in the solution. Molarity of nitrate ions in the solution is the ratio of moles of nitrate ions present in the solution and the volume of solution in liters.

Complete step by step solution
According to the question, $ 100ml $ $ 1M $ $ AgN{O_3} $ is treated with $ 100ml $ $ 1M $ $ NaCl $ , the reaction would be:
 $ AgN{O_3} + NaCl \to NaN{O_3} + AgCl $
In this reaction we can see $ 1mol $ of $ AgN{O_3} $ reacts with $ 1mol $ of $ NaCl $ to give $ 1mol $ of $ NaN{O_3} $ .
Now we have given that the volume of $ 1M $ $ AgN{O_3} $ is $ 100ml $ ,
So, number of moles of $ AgN{O_3} $ = Molarity x Volume of solution in liter
 $ \begin{gathered}
   \Rightarrow 1M \times 0.1L \\
   \Rightarrow 0.1mol \\
\end{gathered} $
We have also given that the volume of $ 1M $ $ NaN{O_3} $ is $ 100ml $ ,
So, number of moles of $ NaN{O_3} $ = Molarity x Volume of solution in liter
  $ \begin{gathered}
   \Rightarrow 1M \times 0.1L \\
   \Rightarrow 0.1mol \\
\end{gathered} $
As we have seen above that $ 1mol $ of $ AgN{O_3} $ reacts with $ 1mol $ of $ NaCl $ to give $ 1mol $ of $ NaN{O_3} $
So, $ 0.1mol $ of $ AgN{O_3} $ reacts with $ 0.1mol $ of $ NaCl $ to give $ 0.1mol $ of $ NaN{O_3} $
So, number of moles of nitrate ions are $ 0.1mol $
Total volume of solution will be: $ (100 + 100)ml \Rightarrow 200ml $
So, molarity of nitrate ions = $ \dfrac{{0.1mol}}{{0.2L}} $
 $ \Rightarrow 0.5M $
Hence, option D is correct.

Note:
There is one more method to determine concentration that is molality. It is represented by $ 'm' $ . To calculate molality we must know the weight of the solvent. As Molality is defined as the number of moles of solute present in $ 1kg $ of solvent. Mathematically, molality is the ratio of number of moles of solute and mass of solvent in kilogram. A commonly used unit for molality is $ mol/kg $ . A solution of concentration $ 1mol/kg $ is also sometimes denoted as $ 1molal $ or $ 1m $ .