12. Determine the molar concentration of ethanol (CH,CH:OH, 46.08 g/mol) in a w
ID: 1049167 • Letter: 1
Question
12. Determine the molar concentration of ethanol (CH,CH:OH, 46.08 g/mol) in a w I) in a wine that is 14% ethanol by mass. The density of this wine is 0.93 g/cm? a. 0.063 M b. 13.0 M c. 0.14 M d. 2.8 M e. 3.0 M 13. A salt solution is added to a marine aquarium. What mass of sodium chloride is needed to produce 250.0 mL of a solution that has a concentration of 50.0 mM? a. 731 g b. 731 mg c. 58.5 g d. 2.92 mg e. 2.92 g 14. Commercial hydrochloric acid is 12.1 M. What volume of commercial HCI solution should be used to prepare 250.0 mL of 3.00 M HCI? d. 62.0 mL e. 83.0 mL a. 139 mL b. 126 mL c. 252 mLExplanation / Answer
The wine is 14% ethanol by mass. this means 14gm ethanol is present in 100gm of wine.
Moles of ethanol present in 100gm of wine =14gm/molar mass = 14gm/46.08gmmol^-1 = 0.304 moles
Density of wine =0.93 gm/c.c. So, volume of 100 gm wine =100gm/0.93 gm/c.c = 107.53 c.c = 107.53 mL
So, 107.53 mL wine contains 0304 mole of ethanol
Molarity = number of moles/1000mL
Molarity of ethanol = 0.304 moles *1000mL/ 107.53 mL = 2.83 M
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50mM = 50*10^-3 M = 0.05 moles in 1000mL
Number of moles present in 250mL solution = 0.05 moles*250/1000mL =0.0125 moles
Mass of NaCl = moles * molar mass =0.0125 moles*58.5gm/mol =731.25 mg
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M1V1 = m2V2
where M1 = 12.1M, V1 = ?, M2 = 3M,V2 = 250mL
V1 = M2V2/M1 = 3M*250mL/12.1 M= 61.98 mL ~62mL
Answer is D
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