StreakPeaked· Practice

ExamsJEE AdvancedChemistry

Match the following (K = 39, Ca = 40, Na = 23, Cl = 35.5): List-I: (P) 20% (w/w) KOH solution, density = 1.02 g/mL (Q) 444 g of CaCl2 dissolved in water to make 1 L solution (R) Volume of 1.204 x 10²⁴ molecules of water at 4 deg C (S) Volume of 0.2 M NaOH solution containing 40 mg of NaOH List-II: (1) 4.00 M (2) 3.64 M (3) 5 mL (4) 36 mL (5) 70 mL

  1. P→1; Q→2; R→3; S→4
  2. P→2; Q→3; R→5; S→1
  3. P→2; Q→1; R→4; S→3
  4. P→1; Q→2; R→4; S→5

Correct answer: P→2; Q→1; R→4; S→3

Solution

(P) M = (10 x 1.02 x 20)/56 = 3.64 M → 2. (Q) moles = 444/111 = 4 mol in 1 L → 4.00 M → 1. (R) moles H2O = 1.204e24/6.02e23 = 2; mass = 36 g; at 4°C, V = 36 mL → 4. (S) moles NaOH = 0.040/40 = 0.001; V = 0.001/0.2 = 0.005 L = 5 mL → 3.

Related JEE Advanced Chemistry questions

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →