JEE Main 2021 · 26 August · evening · Numerical answer · +4 / −0
100 mL of Na₃PO₄ solution contains 3.45 g of sodium. The molarity of the solution is _____________ × 10⁻² mol L⁻¹. (Nearest integer)
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Answer and solution
Answer 50
Solution
Chemistry ma’am · 3 steps
Count moles of sodium first, then remember each formula unit of the salt carries three of them.
- Moles of sodium. n(Na) = (3.45 g)/(23.0 g mol⁻¹) = 0.15 mol.
- Moles of salt. One Na₃PO₄ gives three Na⁺ ions, so the moles of salt are one third of the moles of sodium: n(Na₃PO₄) = 0.15/3 = 0.05 mol.
M = n(Na₃PO₄)/(V (L)) = (0.05 mol)/(0.1 L) = 0.5 mol L⁻¹
- So. 0.5 mol L⁻¹ = 50 × 10⁻² mol L⁻¹, so the answer is 50.
The common slip: 150. Using the 0.15 mol of sodium as the moles of salt gives 1.5 M, which is 150. You need to divide by 3, because each Na₃PO₄ holds three sodium atoms.
The idea to keep. When the mass of one element in a salt is given, convert it to moles of that element. Then divide by its subscript in the formula to get moles of the salt.
Answer from the NTA final answer key. Solution by PrepWiser.