1. A mixture of alcohol and water is 28.00% (v/v) alcohol. How many L of the mix
ID: 708425 • Letter: 1
Question
1. A mixture of alcohol and water is 28.00% (v/v) alcohol. How many L of the mixture do I need in order to have 250.0 mL of alcohol? 2. A mixture of salt and water is 28.0 g salt and 60.0 g water. What is the % (m/m) salt? 3. A salt water mixture is 33.0% (w/w) walt. (The w/w refers to weight / weight or mass / mass). If there are 300.0 g salt in the mixture, what is the total weight of the mixture? 4. A sand-water mixture weights 38.6 g. If the mixture is 15.0% (m/m) water, how many grams of water are in the mixture? 1. A mixture of alcohol and water is 28.00% (v/v) alcohol. How many L of the mixture do I need in order to have 250.0 mL of alcohol? 2. A mixture of salt and water is 28.0 g salt and 60.0 g water. What is the % (m/m) salt? 3. A salt water mixture is 33.0% (w/w) walt. (The w/w refers to weight / weight or mass / mass). If there are 300.0 g salt in the mixture, what is the total weight of the mixture? 4. A sand-water mixture weights 38.6 g. If the mixture is 15.0% (m/m) water, how many grams of water are in the mixture?Explanation / Answer
1.
28% (v/v) means 28 L of solute (alcohol) is dissolved (or added) in 100 L of mixture.
Thus water in 100 L mixture = 100 - 28 = 72 L.
Thus 28 L alcohol needs 72 L water or 1 L alcohol needs 72/28 L water
Therefore 250 mL or 0.250 L of alcohol will need water = 0.250*72/28
= 0.64 L of water
2.
% (m/m) salt = (mass of salt / mass of solution) * 100
= [28/ (28 + 60)]*100 = 32%
3.
33% (w/w) means 33 g of salt is dissolved (or added) in 100 g of mixture.
Thus weight of mixture containing 1 g salt is 100/33 g
Hence weight of mixture containing 300.0 g = 300.0*100/33 g = 909.1 g
4.
15% (w/w) water means 15 g of water is in 100 g of mixture.
Thus water present in 1 g mixture = 15/100 g
Hence water present in 38.6 g mixture = 38.6*15/100 g = 5.79 g
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