A solution is prepared by dissolving 65 g of ethyl alcohol (density=0.78 g/mL) i
ID: 714130 • Letter: A
Question
A solution is prepared by dissolving 65 g of ethyl alcohol (density=0.78 g/mL) in 2.35 L of water. (ethyl alcohol formula: CH3CH2OH). Assume that volumes of both solute and solvent are additive which means there is no net change in volume when the two volumes are mixed. The total volume is simply the sum of individual volumes mixed together. a. Calculate the solute concentration in percent (by volume) units.
b. Calculate the solute concentration in percent (by mass) units.
c. Calculate the solute concentration in molarity units.
d. Calculate the mol% of solute assuming that the solute is ethyl alcohol (CH3CH2OH).
Explanation / Answer
Solution :-
Mass of ethyl alcohol =65 g
Density of ethyl alcohol = 0.78 g/ml
Volume of water = 2.35 L
Volume of ethyl alcohol = mass / density
= 65 g / 0.78 g per ml
= 83.3 ml
Converting ml to liter
83.3 ml * 1 L / 1000 ml = 0.0833 L
#a) lets calculate the volume percent of the solute
Volume % = [volume of solute / total volume ] * 100 %
= [0.0833 L / (2.35 L + 0.0833 L)] * 100 %
= 3.42 %
#b) calculating the solute concentration In percent by mass
Lets first calculate the mass of water
Mass of water = volume x density
Mass of water = (2.35 L * 1000 ml / 1 L ) * 1 g/L
= 2350 g
Mass % of solute = [ mass of solute / ( mass of solute + mass of solvent )] * 100 %
= [65 g / (65 g + 2350 g)]*100 %
= 2.69 % ( by mass )
#c) Calculating the solute concentration in molarity
lets first calculate moles
Moles = mass / molar mass
Molar mass of ethyl alcohol = 46.07 g /mol
Moles of ethyl alcohol = 65 g / 46.07 g per mol
=1.41 mol
Total volume = ethyl alcohol + water
= 0.0833 L + 2.35 L
=2.43 L
Molarity = moles / volume
= 1.41 mol / 2.43 L
= 0.580 M
Therefore molarity is 0.580 M
#d) Calculating the mole percent
Lets first calculate moles of both solute and solvent
We have volume of water which we can convert to mass of water using the density and volume of water
water density is 1 g/ml
Mass of water = (2.35 L* 1000 ml / 1 L) * 1 g per ml = 2350 g water
moles = mass / molar mass
Moles of ethyl alcohol = 65 g / 46.07 g per mol
= 1.41mol
Moles of water = 2350 g / 18.016 g per mol
= 130.4 mol water
Mol % of solute =[ moles of solute /( moles of solute + moles of solvent ) ] * 100 %
= [1.41 mol / (1.41mol + 130.4 mol)] *100 %
= 1.07 %
Therefore mol % of solute is 1.07 %
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