LIST I LIST II (P) 3.42 g C 12 H 22 O 11 in 100 ml solution (1) 0.2 M (Q) Mole fraction of C 2 H 5 OH (aqueous solution) = 0.2 (2) 0.17 m (R) 18 g C 6 H 12 O 6 in 500 ml solution ( d =

Question

LIST I LIST II (P) 3.42 g C 12 ​ H 22 ​ O 11 ​ in 100 ml solution (1) 0.2 M (Q) Mole fraction of C 2 ​ H 5 ​ OH (aqueous solution) = 0.2 (2) 0.17 m (R) 18 g C 6 ​ H 12 ​ O 6 ​ in 500 ml solution ( d = 1.2 g / ml ) (3) 0.05 m (S) 6g urea ( NH 2 ​ CONH 2 ​ ) dissolved in 2 kg water (4) 0.1 M (5) 13.9 m Choose the correct match

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(I) M = 342 3.42 × 10 ​ = 0.1 (II) m = 0.8 × 18 0.2 × 1000 ​ = 13.9 m (III) M = 180 × 0.5 18 ​ = 0.2 M m = 180 × 582 18 × 1000 ​ = 0.17 m (IV) m = 60 × 2 6 ​ = 0.05 m

Chemistry - Concentration Terms Question with Solution | TestHub

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JEE Main (2028)

JEE Main - 1 year (12th class)

LIST I	LIST II
(P) $3.42 g C_{12} H_{22} O_{11}$ in 100 ml solution	(1) 0.2 M
(Q) Mole fraction of $C_{2} H_{5} OH$ (aqueous solution) = 0.2	(2) 0.17 m
(R) $18 g C_{6} H_{12} O_{6}$ in 500 ml solution ( $d = 1.2 g / ml$ )	(3) 0.05 m
(S) 6g urea ( $NH_{2} CONH_{2}$ ) dissolved in 2 kg water	(4) 0.1 M
	(5) 13.9 m