Question
Question: If \( 100ml \) of each of \( 1M \) \( AgN{O_3} \) and \( 1M \) \( NaCl \) are mixed. The nitrate ion...
If 100ml of each of 1M AgNO3 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
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, 100ml 1M AgNO3 is treated with 100ml 1M NaCl , the reaction would be:
AgNO3+NaCl→NaNO3+AgCl
In this reaction we can see 1mol of AgNO3 reacts with 1mol of NaCl to give 1mol of NaNO3 .
Now we have given that the volume of 1M AgNO3 is 100ml ,
So, number of moles of AgNO3 = Molarity x Volume of solution in liter
⇒1M×0.1L ⇒0.1mol
We have also given that the volume of 1M NaNO3 is 100ml ,
So, number of moles of NaNO3 = Molarity x Volume of solution in liter
⇒1M×0.1L ⇒0.1mol
As we have seen above that 1mol of AgNO3 reacts with 1mol of NaCl to give 1mol of NaNO3
So, 0.1mol of AgNO3 reacts with 0.1mol of NaCl to give 0.1mol of NaNO3
So, number of moles of nitrate ions are 0.1mol
Total volume of solution will be: (100+100)ml⇒200ml
So, molarity of nitrate ions = 0.2L0.1mol
⇒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 .