List - I List - II (P) 4.5 m solution of CaCO 3 , density 1.45 gm / ml (1) mole fraction of solute is 0.2 (Q) 3 M of 100 ml H 2 SO 4 mixed with 1 M of 300 ml H 2 SO 4 solution (2) mass of th

Question

List - I List - II (P) 4.5 m solution of CaCO 3 ​ , density 1.45 gm / ml (1) mole fraction of solute is 0.2 (Q) 3 M of 100 ml H 2 ​ SO 4 ​ mixed with 1 M of 300 ml H 2 ​ SO 4 ​ solution (2) mass of the solute is 360 gm (R) 14.5 m solution of CaSO 4 ​ (3) molarity = 4.5 (S) in 4M of 2 litre solution of NaOH, 40 gm NaOH is added. (4) molarity = 1.5 (T) 5 m (molal) NaOH solution (5) 16.66% ( w / w ) of NaOH in solution

TestHub · Accepted Answer

P: Molarity = 4.5 M (3)

Q: Resultant molarity = $(3 \times 100 + 1 \times 300) /400 = 1.5$ M (4)

R: Mole fraction = $14.5/ (14.5 + 55.5) = 0.2$ (1)

S: Total moles NaOH = $4 \times 2 + 40/40 = 9$ moles. Mass = $9 \times 40 = 360$ g (2). Molarity = $9/2 = 4.5$ M (3)

T: $%$ (w/w) NaOH = $(5 \times 40) / (5 \times 40 + 1000) \times 100 = 16.66%$ (5)

Chemistry - Concentration Terms Question with Solution | TestHub

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Doubts & Discussion

JEE Main (2028)

List - I	List - II
(P) 4.5 m solution of $CaCO_{3}$ , density $1.45 gm / ml$	(1) mole fraction of solute is 0.2
(Q) 3 M of $100 ml H_{2} SO_{4}$ mixed with 1 M of $300 ml H_{2} SO_{4}$ solution	(2) mass of the solute is 360 gm
(R) 14.5 m solution of $CaSO_{4}$	(3) molarity = 4.5
(S) in 4M of 2 litre solution of NaOH, 40 gm NaOH is added.	(4) molarity = 1.5
(T) 5 m (molal) NaOH solution	(5) $16.66% (w / w)$ of NaOH in solution