Match column-I with column-II
column-I column-II
A 200 ml solution containing 1.06 gm of Na₂CO₃(Mol.Wt=106) I mole fraction of solvent is 0.2
B 100 ml of solution containing 4.9 gm of H₃PO₄(Mol.Wt=98) II mole fraction of solute is 0.2
C 252 gm aqueous solution containing 1 mole of glucose(Mol.Wt=180) III deci normal solution
D 0.06 gms of urea(Mol.Wt=60) added to 100 ml water(density = 1gm/ml) IV semi molar solution
V centi molal solution

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A	B	C	D
III	V	II	IV

A	B	C	D
III	II	IV	V

A	B	C	D
II	IV	III	V

A	B	C	D
III	IV	II	V

answer is A.

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Detailed Solution

	column-I		column-II
A	200 ml solution containing 1.06 gm of Na₂CO₃(Mol.Wt=106)	III	$\frac{1.06}{53} \times \frac{1000}{200} = 0.1$ (deci normal solution)
B	100 ml of solution containing 4.9 gm of H₃PO₄(Mol.Wt=98)	IV	$\frac{4.9}{98} \times \frac{1000}{100} = 0.5$ (semi molar solution)
C	252 gm aqueous solution containing 1 mole of glucose(Mol.Wt=180)	II	Mass of water = 252-180=72 mole fraction of glucose = $\frac{1}{1 + 4} = 0.2$ (mole fraction of solute is 0.2)
D	0.06 gms of urea(Mol.Wt=60) added to 100 ml water(density = 1gm/ml)	V	$\frac{0.06}{60} \times \frac{1000}{100} = 0.01$ (centi molal solution)