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Question: Concentrated aqueous sulphuric acid is \(98%\) \({{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\) by mass ...

Concentrated aqueous sulphuric acid is 9898% H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}} by mass and has a density of 1⋅84 gm/ml1\cdot 84\text{ gm/ml}. What volume of the concentrated acid is required to make 5⋅0 L5\cdot 0\text{ L} of 0⋅50 M0\cdot 50\text{ M} H2SO4{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}} solution?
(A) 250 ml
(B) 135⋅86 ml\text{135}\cdot \text{86 ml}
(C) 460 ml
(D) 245 ml

Explanation

Solution

Molarity can be defined as the number of moles of a solute present in one liter of solution. Concentration of a solution can be calculated by using the molarity formula. Molarity is also known as molar concentration of a solution. Thus, we can say that the quantitative measure of molarity and concentration are the same.

Complete Step by step solution:
First, we have to calculate the number of moles present in the given solution
Number of moles =Given massMass of solute=9898=1 ml\text{Number of moles =}\dfrac{\text{Given mass}}{\text{Mass of solute}}=\dfrac{98}{98}=1\text{ ml}
Given that,
Density of the solution =1.84 gm/ml=1.84\text{ gm/ml}
Now, in order to calculate volume, we know that
Density =MV=MassVolume\text{Density =}\dfrac{\text{M}}{\text{V}}=\dfrac{\text{Mass}}{\text{Volume}}
Thus,
Volume =MassDensity\text{Volume =}\dfrac{\text{Mass}}{\text{Density}}
=100 g1⋅84 g/ml=\dfrac{100\text{ g}}{\text{1}\cdot \text{84 g/ml}}
=54⋅34 ml=54\cdot 34\text{ ml}
Now, to calculate the molarity, we know that
Molarity = Mole of soluteVolume of solution in ml×1000\text{Molarity = }\dfrac{\text{Mole of solute}}{\text{Volume of solution in ml}}\times 1000
Molarity = 154⋅34×1000\text{Molarity = }\dfrac{1}{54\cdot 34}\times 1000
Molarity = 18⋅40 M\text{Molarity = 18}\cdot \text{40 M}
According to law of neutralization
M1V1 = M2V2{{\text{M}}_{1}}{{\text{V}}_{1}}\text{ = }{{\text{M}}_{2}}{{\text{V}}_{2}}
18⋅40×V1=0⋅5×5⋅0×100018\cdot 40\times {{\text{V}}_{1}}=0\cdot 5\times 5\cdot 0\times 1000
V1=0⋅5×0⋅5×100018⋅40{{\text{V}}_{1}}=\dfrac{0\cdot 5\times 0\cdot 5\times 1000}{18\cdot 40}
V1=250018⋅40{{\text{V}}_{1}}=\dfrac{2500}{18\cdot 40}
V1=135⋅86 ml{{\text{V}}_{1}}=135\cdot 86\text{ ml}

Additional information:
Density: Density of a solution can be defined as the mass per unit volume. The density of a solution is represented by the ρ\rho . The formula of density is ρ=MassVolume\rho =\dfrac{\text{Mass}}{\text{Volume}}.
Molality: Molality of a solution can be defined as the amount of a substance which is dissolved in a definite mass of a solvent. In addition to this, it is also defined as the moles of a solute per kilogram of a solvent. The unit of molality is m or mol/kgm\text{ or mol/kg}.

Note: For these types of questions, we can also use the concept of normality. Normality can be defined as the number of gram equivalents of solute present in one liter of solution. In addition to this, normality of a solution is represented by N\text{N}.