JEE Main 2020 · 4 September · evening · Numerical answer · +4 / −0
A 100 mL solution was made by adding 1.43 g of Na₂CO₃.xH₂O. The normality of the solution is 0.1 N. The value of x is _____.
(The atomic mass of Na is 23 g/mol)
Answer and solution
Answer 10
Solution
Chemistry ma’am · 4 steps
Turn normality into moles using the n-factor of 2, then the molar mass tells you how much water is in there.
- Find the n-factor. Na₂CO₃ gives CO₃²⁻, which takes up 2 H⁺ ions. So its n-factor is 2, and molarity = normality/2 = 0.1/2 = 0.05 M.
- Moles in solution. The volume is 100 mL = 0.1 L, so moles of salt = 0.05 × 0.1 = 0.005 mol.
M = (1.43 g)/(0.005 mol) = 286 g mol⁻¹
- Split the molar mass. Na₂CO₃ alone is 46 + 12 + 48 = 106. The water part is 286 - 106 = 180, and 180 ÷ 18 = 10.
- So. The salt is Na₂CO₃.10H₂O, and x = 10.
The common slip: 2. Taking the n-factor of Na₂CO₃ as 1 gives 0.01 mol and a molar mass of 143, so x comes out near 2 instead of 10.
The idea to keep. Normality counts equivalents, not moles. Divide by the n-factor to get moles, and then the hydrate's molar mass minus the anhydrous part gives the water.
Answer from the NTA final answer key. Solution by PrepWiser.