Question

The molar concentration of the chloride ion in the solution obtained by mixing $300mL$ of $3.0M$ $NaCl$ and $200mL$ of $4.0M$ solution of $BaC{l_2}$ is:
A: $1.6M$
B: $1.8M$
C: $5.0M$
D: $0.5M$

Answer 1

Hint: Molarity is defined as the number of moles of the substance per liter of the solution and normality is defined as the number of gram equivalents of the substance dissolved per liter of the solution. Gram equivalents can be found by multiplying valency with number of moles of substance.
Formula used: ${N_1}{V_1} + {N_2}{V_2} = NV$
$V = {V_1} + {V_2}$
Normality$ = $molarity$ \times n - $factor
Where ${N_1},{V_1}$ is the normality and volume of $NaCl$ solution and ${N_2},{V_2}$ is the normality and volume of $BaC{l_2}$ solution.

Complete step by step answer:
In this question $NaCl$ solution and $BaC{l_2}$ solution are mixed together and we have to find the molarity of the resulting chloride ions. $NaCl$ solution gives only one chloride ion this means $n - $factor of $NaCl$ is one and $BaC{l_2}$ solution gives two chloride ion this means $n - $factor of $BaC{l_2}$ is two. With the help of this and molarity (given in question) we can find the normality of these solutions.
Molarity of given $NaCl$ solution is $3.0M$ and $n - $factor of the solution is one. Therefore we can calculate normality of $NaCl$ solution with the formula:
Normality$ = $molarity$ \times n - $factor
Normality of $NaCl$ solution$ = 3 \times 1 = 3$
Similarly, the molarity of given $BaC{l_2}$ solution is $4.0M$ and $n - $factor of the solution is two. Therefore we can calculate normality of $BaC{l_2}$ solution with the formula:
Normality$ = $molarity$ \times n - $factor
Normality of $BaC{l_2}$ solution$ = 4 \times 2 = 8$
Total volume of the solution obtained by mixing the solution of $BaC{l_2}$ and $NaCl$ will be equal to the sum of these solutions. Volume of $NaCl$ solution is $300mL$ and the volume of $BaC{l_2}$ solution is $200mL$. So, total volume of the solution $\left( V \right)$ will be:
$V = 300 + 200 = 500mL$
Now, we have obtained all the quantities required to calculate normality of chloride ions of resulting solution.
$V = 500mL$
Volume of $NaCl$ solution$ = {V_1} = 300mL$
Volume of $BaC{l_2}$ solution$ = {V_2} = 200mL$
Normality of $NaCl$ solution$ = {N_1} = 3N$
Normality of $BaC{l_2}$ solution$ = {N_2} = 8N$
Using the formula written above we can calculate normality of the resulting solution. That is,
${N_1}{V_1} + {N_2}{V_2} = NV$
$\left( {3 \times 300} \right) + \left( {8 \times 200} \right) = N \times 500$
Solving this equation we get,
$N \times 500 = 900 + 1600 = 2500$
$N = 5$
So, the normality of chloride ions in the resulting solution is $5N$. Normality of chloride ions is equal to the molarity of chloride ions as charge on chloride ions is one so, their $n - $factor is one. Therefore molarity of chloride ions of the resulting solution is $5M$.
So, the correct answer is option C.

Note:
Molality of a solution is defined as the number of moles of the solute that are present in one kilogram of solvent. In calculating molarity volume of the solution is used but for calculating molality mass of solvent is used.