ANS:D - 0.5055

To find the composition of the aqueous solution expressed in normality, we first need to calculate the molarity of the solution and then convert it to normality since normality is based on equivalents per liter rather than moles per liter. Given:

• Specific gravity of the solution = 1.011
• Weight percentage of 𝐻2𝑆𝑂4H2​SO4​ in the solution = 2.45%

First, let's calculate the molarity of the solution:

1. Calculate the density of the solution in g/mL: Since specific gravity is the ratio of the density of the solution to the density of water, and the density of water is 1 g/mL, the density of the solution is 1.011×1 g/mL=1.011 g/mL1.011×1g/mL=1.011g/mL.
2. Calculate the mass of 𝐻2𝑆𝑂4H2​SO4​ in 100 mL of solution: 2.45 g2.45g of 𝐻2𝑆𝑂4H2​SO4​ per 100 g of solution
3. Calculate the molar mass of 𝐻2𝑆𝑂4H2​SO4​: 𝐻2𝑆𝑂4H2​SO4​ molar mass = (2×1)+32+(4×16)=98 g/mol(2×1)+32+(4×16)=98g/mol
4. Calculate the number of moles of 𝐻2𝑆𝑂4H2​SO4​ in 100 mL of solution: moles=massmolar mass=2.45 g98 g/molmoles=molar massmass​=98g/mol2.45g​
5. Calculate the molarity of the solution: Since 1000 mL = 1 L, the molarity (𝑀M) is given by: 𝑀=molesvolume in liters=moles0.1 LM=volume in litersmoles​=0.1Lmoles​

Now, to convert molarity to normality, we consider the acid-base reaction of sulfuric acid (𝐻2𝑆𝑂4H2​SO4​): 𝐻2𝑆𝑂4→2𝐻++𝑆𝑂42−H2​SO4​→2H++SO42−​ Since each mole of 𝐻2𝑆𝑂4H2​SO4​ produces 2 moles of 𝐻+H+, the normality (𝑁N) is twice the molarity (𝑀M). 𝑁=2𝑀N=2M Let's calculate: 𝑀=2.45 g98 g/mol×0.1 L=0.25 mol/LM=98g/mol×0.1L2.45g​=0.25mol/L 𝑁=2×0.25 mol/L=0.50 NN=2×0.25mol/L=0.50N Therefore, the composition of the aqueous solution expressed in normality is 0.50 N0.50N. So, the correct answer is 0.5000 N0.5000N.