The molecular mass of sulphuric acid is 98. If 49 g of pure H2SO4 is dissolved in water to prepare one litre of solution, what will be the normality (strength) of the acid solution?

Introduction / Context:
This question tests your understanding of the concept of normality in solution chemistry and how to calculate it for an acid like sulphuric acid. In titration and concentration calculations, it is important to know not only molarity but also normality, especially when dealing with acids and bases that can donate or accept more than one proton. Sulphuric acid, H2SO4, is a diprotic acid, so each mole can supply two equivalents of hydrogen ions in reactions where full neutralisation occurs.

Given Data / Assumptions:

• Molecular mass of H2SO4 is given as 98 g per mole.
• Mass of H2SO4 used = 49 g.
• Volume of the solution prepared = 1 litre.
• We assume complete ionisation to give two hydrogen ions per molecule in the context of normality.

Concept / Approach:
Normality (N) is defined as the number of gram equivalents of solute per litre of solution. For an acid, gram equivalent mass is given by molecular mass divided by basicity, where basicity is the number of ionisable hydrogen ions per molecule. Sulphuric acid has basicity 2, because each molecule can provide two H+ ions in neutralisation reactions. Therefore, equivalent mass of H2SO4 = 98 / 2 = 49 g. One gram equivalent of H2SO4 is thus 49 g. If 49 g is dissolved to make one litre, the solution is 1 equivalent per litre, that is, 1 N.

Step-by-Step Solution:
Step 1: Note the molecular mass of H2SO4, which is 98 g per mole. Step 2: Identify the basicity of sulphuric acid. H2SO4 can release two hydrogen ions per molecule in neutralisation, so its basicity is 2. Step 3: Calculate the equivalent mass of H2SO4 using the formula: equivalent mass = molecular mass / basicity. Step 4: Substitute values: equivalent mass = 98 / 2 = 49 g per equivalent. Step 5: The problem states that 49 g of H2SO4 is dissolved to make 1 litre of solution. Step 6: Since 49 g is exactly one equivalent mass, the solution contains 1 equivalent of H2SO4 in 1 litre. Step 7: Normality N is defined as equivalents per litre, so N = 1 equivalent per litre = 1 N.

Verification / Alternative check:
You can also think in terms of molarity and then convert to normality. If 49 g of H2SO4 is present, that is 49 / 98 = 0.5 moles in one litre, so molarity is 0.5 M. For a diprotic acid that donates two hydrogen ions, normality for acid base reactions is N = basicity * molarity = 2 * 0.5 = 1 N. This alternative route leads to the same answer and confirms the calculation.

Why Other Options Are Wrong:
Two normal would require 2 equivalents per litre, which for H2SO4 would be 2 * 49 = 98 g per litre, not 49 g. A value of 0.5 normal would correspond to 0.5 equivalents, or 24.5 g of H2SO4 per litre. Four normal would require 4 equivalents, that is, 196 g per litre. A value of 0.25 normal would require only 12.25 g per litre. None of these match the given 49 g per litre situation.

Common Pitfalls:
Students often confuse molarity with normality and may answer 0.5, thinking in terms of moles only. It is crucial to adjust for the number of ionisable hydrogen ions in acids or hydroxyl groups in bases. Another common error is to misapply the basicity, using 1 instead of 2 for H2SO4. Always check whether the acid is monoprotic, diprotic or triprotic before computing equivalent mass or normality.

Final Answer:
The normality of the solution prepared by dissolving 49 g of H2SO4 in one litre of solution is one normal (1 N).