In a tank 100 liters of brine solution containing 50 kg of total dissolved salt. Pure water is allowed to run into the tank at the rate of 3 L/min and brine solution runs out at the rate of 2 L/min. The solution is kept stirred to make the concentration uniform. Compute the kg of salt in the tank after 1.0 hour.Question 6Select one:A.13.59 kgB.19.35 kgC.19.53 kgD.15.93 kg

To solve this problem, we need to calculate the amount of salt in the tank after 1.0 hour.

First, let's calculate the amount of water that enters the tank in 1.0 hour. Since water is flowing into the tank at a rate of 3 L/min, in 1.0 hour (60 minutes), the amount of water entering the tank is 3 L/min * 60 min = 180 L.

Next, let's calculate the amount of brine solution that leaves the tank in 1.0 hour. Since brine solution is flowing out at a rate of 2 L/min, in 1.0 hour (60 minutes), the amount of brine solution leaving the tank is 2 L/min * 60 min = 120 L.

Now, let's calculate the total volume of the solution in the tank after 1.0 hour. The initial volume of the solution in the tank is 100 L. Since 180 L of water enters and 120 L of brine solution leaves, the total volume of the solution in the tank after 1.0 hour is 100 L + 180 L - 120 L = 160 L.

To find the concentration of salt in the tank after 1.0 hour, we can use the formula:

Concentration of salt = (mass of salt) / (volume of solution)

The initial concentration of salt in the tank is 50 kg / 100 L = 0.5 kg/L.

Since the solution is kept stirred to make the concentration uniform, the concentration of salt remains the same throughout the tank.

Therefore, the mass of salt in the tank after 1.0 hour is:

Mass of salt = Concentration of salt * Volume of solution Mass of salt = 0.5 kg/L * 160 L = 80 kg

Therefore, the kg of salt in the tank after 1.0 hour is 80 kg.

So, the correct answer is not provided in the options.