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Question: Given that:\[25{\text{ }}ml{\text{ }}of{\text{ }}3.0M{\text{ }}HN{O_3}\] are mixed with \[75{\text{ ...

Given that:25 ml of 3.0M HNO325{\text{ }}ml{\text{ }}of{\text{ }}3.0M{\text{ }}HN{O_3} are mixed with 75 ml of 4.0M HNO375{\text{ }}ml{\text{ }}of{\text{ }}4.0M{\text{ }}HN{O_3} . If the volumes are additive, the molarity of the final mixture would be:

B)\;\;\;4.0M C)\;\;3.75M D)\;\;3.5M$$
Explanation

Solution

Homogeneous mixtures or solutions are the one in which mixtures are of uniform composition. Heterogeneous mixtures are the ones with non-uniform composition. Solvents are the chemicals in mixture which are present in the largest amount and solutes are the other components. When we talk about Molarity or molar concentration we must know that it is the number of moles of solute per litre of solution.

Complete step-by-step answer: Molar concentration can be used to convert between the mass or moles of solute and the volume of the solution. Moles/litreMoles/litre is one of the most common units used to measure the concentration of a solution. Molarity is often used to calculate the volume of solvent or amount of solute. If we wish to represent the relationship between two solutions with same amount of moles of solute we can do it by the formula c1V1= c2V2{c_1}{V_1} = {\text{ }}{c_2}{V_2} where   c\;c is concentration and VV is volume.
The molarity of mixing is a very important concept. In that we need to understand that there are three samples of solution which have same solvent and solute and their molarity being M1, M2, M3{M_1},{\text{ }}{M_2},{\text{ }}{M_3} and volumes V1, V2, V3{V_1},{\text{ }}{V_2},{\text{ }}{V_3} respectively. These solutions are mixed; molarity of mixed solution may be given as: M1V1 + M2V2 + M3V3 = MR(V1 + V2 + V3){M_1}{V_1}{\text{ }} + {\text{ }}{M_2}{V_2}{\text{ }} + {\text{ }}{M_3}{V_3}{\text{ }} = {\text{ }}MR\left( {{V_1}{\text{ }} + {\text{ }}{V_2}{\text{ }} + {\text{ }}{V_3}} \right) Where MRMR= resultant molarity (V1 + V2 + V3)\left( {{V_1}{\text{ }} + {\text{ }}{V_2}{\text{ }} + {\text{ }}{V_3}} \right) = resultant volume

We could easily solve this problem using dilution equation:
M1V1+M2V2=MV{M_1} {V_1} + {M_2 }{V_2} = MV
Where, MMand VV are the molarity and volume of the mixture.
Thus we put the values in the equation such that:
Now after putting the designated values in there places we will get:
3.0(25) + 4.0(75) = M(25+75)3.0\left( {25} \right){\text{ }} + {\text{ }}4.0\left( {75} \right){\text{ }} = {\text{ }}M\left( {25 + 75} \right)
Now to find MM we will further go as:
375 = M(100)375{\text{ }} = {\text{ }}M\left( {100} \right)
Thus MM will come out to be:
M = 3.75M{\text{ }} = {\text{ }}3.75 option ( C )\left( {{\text{ }}C{\text{ }}} \right)
So, the molarity of the mixture solution is 3.75M3.75M .

Note: We must know that dilution is the process by which a solution is made less concentrated by addition of more solvent. The SISI unit for molar concentration is mol/m3mol/{m^3}. A solution that contains one mole of solute per 1 litre1{\text{ }}litre of solution (1 mol/L)\left( {1{\text{ }}mol/L} \right) is called “one Molar” or1 M1{\text{ }}M. The unit mol/Lmol/L can be converted to mol/m3  mol/{m^3}\; by the help of the following equation:
1 mol/L = 1 mol/dm3  = 1 mol dm−3  = 1 M = 1000 mol/m31{\text{ }}mol/L{\text{ }} = {\text{ }}1{\text{ }}mol/d{m^3}\; = {\text{ }}1{\text{ }}mol{\text{ }}d{m^{ - 3}}\; = {\text{ }}1{\text{ }}M{\text{ }} = {\text{ }}1000{\text{ }}mol/{m^3}