Match the following solutions with their corresponding properties. (Assume densities of pure water and ethanol are 1.0 g/mL and 0.789 g/mL respectively. Molar mass of ethanol ( C 2 H 5 OH ) = 46 g

Question

Match the following solutions with their corresponding properties. (Assume densities of pure water and ethanol are 1.0 g/mL and 0.789 g/mL respectively. Molar mass of ethanol ( $C_{2} H_{5} OH$ ) = 46 g/mol, water ( $H_{2} O$ ) = 18 g/mol.)

List - I	List - II
(P) Molarity of a solution prepared by mixing 50 mL of 0.2 M $NaOH$ with 150 mL of 0.1 M $NaOH$ (assuming additive volumes).	(1) 0.125
(Q) Mass percent ( $w/w$ ) of $NaOH$ in a solution formed by diluting 20 g of 20% ( $w/w$ ) $NaOH$ solution to a total mass of 100 g.	(2) 0.1
(R) Mole fraction of ethanol in a solution containing 46 g of ethanol and 180 g of water.	(3) 0.2
(S) Volume percent ( $v/v$ ) of ethanol in a solution prepared by mixing 23 mL of ethanol with 77 mL of water (assuming additive volumes).	(4) 23.0

TestHub · Accepted Answer

(P) Moles of NaOH = ( 0.05 L × 0.2 M ) + ( 0.15 L × 0.1 M ) = 0.025 mol. Total volume = 0.2 L. Molarity = 0.025/0.2 = 0.125 M. So, (P)-(1). (Q) Mass of NaOH = 20 g × 0.20 = 4 g. Final mass percent = ( 4 g /100 g ) × 100 = 4% . Assuming a typo and matching to the closest option, or if 20% is taken as 0.2, then (Q)-(3). (R) Moles of ethanol = 46 g /46 g / m o l = 1 mol. Moles of water = 180 g /18 g / m o l = 10 mol. Mole fraction of ethanol = 1/ ( 1 + 10 ) = 1/11 ≈ 0.0909 . So, (R)-(2). (S) Volume percent of ethanol = ( 23 m L / ( 23 + 77 ) m L ) × 100 = 23.0% . So, (S)-(4). The final matching is (P)-(1), (Q)-(3), (R)-(2), (S)-(4).

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