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Question: If \( 100ml \) of each of \( 1M \) \( AgN{O_3} \) and \( 1M \) \( NaCl \) are mixed. The nitrate ion...

If 100ml100ml of each of 1M1M AgNO3AgN{O_3} and 1M1M NaClNaCl are mixed. The nitrate ion concentration in the resulting solution is:
A. 1M1M
B. 0.25M0.25M
C. 0.75M0.75M
D. 0.5M0.5M

Explanation

Solution

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, 100ml100ml 1M1M AgNO3AgN{O_3} is treated with 100ml100ml 1M1M NaClNaCl , the reaction would be:
AgNO3+NaClNaNO3+AgClAgN{O_3} + NaCl \to NaN{O_3} + AgCl
In this reaction we can see 1mol1mol of AgNO3AgN{O_3} reacts with 1mol1mol of NaClNaCl to give 1mol1mol of NaNO3NaN{O_3} .
Now we have given that the volume of 1M1M AgNO3AgN{O_3} is 100ml100ml ,
So, number of moles of AgNO3AgN{O_3} = Molarity x Volume of solution in liter
1M×0.1L 0.1mol \begin{gathered} \Rightarrow 1M \times 0.1L \\\ \Rightarrow 0.1mol \\\ \end{gathered}
We have also given that the volume of 1M1M NaNO3NaN{O_3} is 100ml100ml ,
So, number of moles of NaNO3NaN{O_3} = Molarity x Volume of solution in liter
1M×0.1L 0.1mol \begin{gathered} \Rightarrow 1M \times 0.1L \\\ \Rightarrow 0.1mol \\\ \end{gathered}
As we have seen above that 1mol1mol of AgNO3AgN{O_3} reacts with 1mol1mol of NaClNaCl to give 1mol1mol of NaNO3NaN{O_3}
So, 0.1mol0.1mol of AgNO3AgN{O_3} reacts with 0.1mol0.1mol of NaClNaCl to give 0.1mol0.1mol of NaNO3NaN{O_3}
So, number of moles of nitrate ions are 0.1mol0.1mol
Total volume of solution will be: (100+100)ml200ml(100 + 100)ml \Rightarrow 200ml
So, molarity of nitrate ions = 0.1mol0.2L\dfrac{{0.1mol}}{{0.2L}}
0.5M\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'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 1kg1kg 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/kgmol/kg . A solution of concentration 1mol/kg1mol/kg is also sometimes denoted as 1molal1molal or 1m1m .