Exams › JEE Main › Chemistry

500 mL of 0.90 M acetic acid (CH3COOH) is combined with 600 mL of 12% w/v acetic acid. Calculate the resulting molarity of acetic acid in the mixed solution. (Molar mass of CH3COOH = 60 g/mol)

1. 0.90 M
2. 1.20 M
3. 1.50 M
4. 1.80 M

Correct answer: 1.50 M

Solution

Moles = 0.45 + 1.20 = 1.65 in total volume 1.1 L, giving 1.5 M.

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

⚔️ Practice JEE Main Chemistry free + battle 1v1 →