Solveeit Logo

Question

Question: 1 M and \(2.5{\text{ L }}NaOH\) solution mixed with another 0.5 M and \(3{\text{ L }}NaOH\) solution...

1 M and 2.5 L NaOH2.5{\text{ L }}NaOH solution mixed with another 0.5 M and 3 L NaOH3{\text{ L }}NaOH solution. Then find out molarity of resultant solution:
(A) 0.80M
(B) 1.0M
(C) 0.73M
(D) 0.50M

Explanation

Solution

The above question can be solved using the formula of molarity.
That is,
Molarity = No. of moles of soluteVolume of solution{\text{Molarity = }}\dfrac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Volume of solution}}}}
Molarity is defined as the total number of moles of solute per liter of solution.
It is represented by M that is molar.
One molar is the molarity of a solution where one gram of solute is dissolved in a liter of solution.
Equation for calculating molarity is,
M=nVM = \dfrac{n}{V}
MM \to Molarity
nn \to No. of moles of solute
VV \to Volume of solution (in liters)

Complete step by step answer:
Now, for calculating the molarity of the resultant solution that is prepared by mixing two solutions.
We will be using the following equation.
M1V1=M2V2{M_1}{V_1} = {M_2}{V_2}
For 1st solution,
M1=1M{M_1} = 1M
V1=2.5L{V_1} = 2.5L
M1=n1V1{M_1} = \dfrac{{{n_1}}}{{{V_1}}}
n1=1×2.5{n_1} = 1 \times 2.5
n1=2.5 moles{n_1} = 2.5{\text{ moles}}
For 2nd solution,
M2=0.5M{M_2} = 0.5M
V2=3L{V_2} = 3L
M2=n2V2{M_2} = \dfrac{{{n_2}}}{{{V_2}}}
n2=0.5×3{n_2} = 0.5 \times 3
n2=1.5 moles{n_2} = 1.5{\text{ moles}}
Total moles of NaOHNaOH in solution
n=n1+n2n = {n_1} + {n_2}
n=2.5+1.5=4.0 molesn = 2.5 + 1.5 = 4.0{\text{ moles}}
So, using the above information.
Total Molarity, M=Total MolesTotal VolumeM = \dfrac{{{\text{Total Moles}}}}{{{\text{Total Volume}}}}
Total=2.5{\text{Total}} = 2.5
Volume+3=5.5L{\text{Volume}} + 3 = 5.5L
M=4.05.5M = \dfrac{{4.0}}{{5.5}}
M=0.73MM = 0.73M

So, the correct answer is Option C.

Note:
During mixing of same mixtures with different concentrations, we must remember it is the number of moles and the total quantity that remain conserved, which further are used to calculate the resultant molarity.