Exams › JEE Main › Chemistry

What amount of magnesium phosphate, Mg3(PO4)2, in moles is needed to provide 0.25 mole of oxygen atoms?

1. 1.25 × 10^−2
2. 2.5 × 10^−2
3. 0.02
4. 3.125 × 10^−2

Correct answer: 3.125 × 10^−2

Solution

Each Mg3(PO4)2 contains 8 O atoms, so moles = 0.25/8 = 0.03125 = 3.125e-2. Stored 2.5e-2 is wrong.

Related JEE Main Chemistry questions

• A 100 mL sample of a urea solution contains 6.02 × 10²⁰ molecules of urea. What is the molarity of the solution? (Take Avogadro constant, N_A = 6.02 × 10²³ mol^−1.)
• When 4.9 g of H2SO4 reacts with NaCl, the products formed are 6 g of sodium hydrogen sulphate and 1.825 g of HCl. What mass of NaCl was decomposed?
• Which of the following has the least mass?
• Haemoglobin has 0.33% iron by mass. If its molecular mass is about 67200 and the atomic mass of iron is 56, how many iron atoms are present in a single haemoglobin molecule?
• For the reaction 2Al(s) + 6HCl(aq) → 2Al³+(aq) + 6Cl^−(aq) + 3H₂(g), which statement is correct?
• Commercial concentrated sulphuric acid contains 95% H₂SO₄ by mass. If its density is 1.834 g cm⁻³, what is the molarity of the solution?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →