Chemistry - Concentration Terms Question with Solution | TestHub
Match the list (Relative atomic mass : Na = 23, K = 39, Ca = 40) :
List - I | List - II |
|---|---|
(I) solution of KOH (density of solution = 1.02 gm/mL) | (P) 4.00 M |
(II) Solution containing in a 1 L solution | (Q) 3.64 M |
(III) Volume of molecules of water at | (R) 5 mL |
(IV) Volume of 0.2 M NaOH solution containing 40 mg of NaOH | (S) 36 mL |
The correct match is :
Options:
Answer:
Solution:
(I) 20% (w/w) KOH: 20 g KOH is present in 100 g of solution. Volume of solution = 100/1.02 = 98.04 mL = 98.04/1000 L. Moles of KOH = 20/56. Molarity = (20/56)/(98.04/1000) = 3.64 M. So I → Q.
(II) 444 g CaCl₂ in 1 litre of solution: moles = 444/111 = 4. Molarity = 4/1 = 4 M. So II → P.
(III) 1.204 × 10²⁴ molecules of water = 2 mol = 36 g. At 4 °C the density of water is 1 g/mL, so the volume is 36 mL. So III → S.
(IV) 40 mg NaOH = 40 × 10⁻³ g / 40 g mol⁻¹ = 10⁻³ mol. Volume of 0.2 M solution = 10⁻³/0.2 = 5 × 10⁻³ L = 5 mL. So IV → R.
Correct match: I-Q, II-P, III-S, IV-R, option (D).
