NEETChemistrySolutions
Match List-I with List-II for an aqueous solution of NaOH which is 20 % (w/w) and has a density of 1.25 g/mL . [Given: Molar mass of NaOH = 40 g mol ⁻¹ ] List-I List-II (A) Molarity (I) 1000 g (B) Molality (II) 250 g (C) Mass of solvent in 1 L solution (III) 6.25 mol L ⁻¹ (D) Mass of solute in 1 L solution (IV) 6.25 mol kg ⁻¹ Choose the correct answer from the options given below:
Options
- A(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
- D(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
Correct answer
B. (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Step-by-step solution
Let us consider 1 L ( 1000 mL ) of the given aqueous NaOH solution. Mass of 1 L solution = Volume Density = 1000 mL 1.25 g/mL = 1250 g . The solution is 20 % (w/w) , which means 20 % of the total mass is the solute ( NaOH ). Mass of solute ( NaOH ) in 1 L solution = 20 100 1250 g = 250 g . This matches (D) with (II). Mass of solvent in 1 L solution = Mass of solution - Mass of solute Mass of solvent = 1250 g - 250 g = 1000 g ( 1 kg ). This matches (C) with (I). Number of moles of NaOH in 1 L solution = Mass Molar m