A certain tank is filled to one quarter of its capacity with a mixture consisting of water and sodium chloride. The proportion of sodium chloride in the tank is 40% by volume and the capacity of the tank is 24 gallons. If the water evaporates from the tank at the rate of 0.5 gallons per hour, and the amount of sodium chloride stays the same, what will be the concentration of water in the mixture in 2 hours?

52 %

Explanation:

The tank is filled to one quarter its capacity

tank capacity filled with mixture = 1/4 * 24 gallons = 6 gallons

percentage of water in the mixture = 100 - 40 = 60%

60 % of 6 gallons = 3.6 gallon of water present

after two hours

the amount lost to evaporation = 0.5 gallons * 2 = 1 gallon

quantity of water remaining = 3.6 - 1 = 2.6 gallon

quantity of the mixture remaining after 1 gallon of water has been removed = 6 - 1 = 5 gallons

the concentration of water in the mixture in percentage = 2.6 / 5 * 100 = 52 %

Answer: C = 52%

Therefore, the final concentration of the mixture is 52%

Explanation:

Given;

The total capacity of the tank = 24 gallons

Rate of evaporation of water r = 0.5 gallons/hour

Time of evaporation t = 2 hours

Initial Volume of mixture in the tank V = 1/4 * 24 = 6 gallons

Initial concentration of the mixture = 40%

Volume of sodium chloride in the mixture will be given as m.

m = 40% V = 40% of 6 = 2.4 gallons

Due to the evaporation the volume of the mixture would have reduced by to V'

V' = V - rt

V' = 6 - 0.5*2

V' = 5 gallons

Since the volume of sodium chloride in the mixture is not affected by the evaporation, m remain 2.4 gallons

the final concentration of water is

C = mass/volume

C = (5.0 - 2.4) / 5

C = 2.6/5

C = 52%

Therefore, the final concentration of the mixture is 52%