Answer:
(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L
Step-by-step explanation:
(a) Initial amount of salt in tank
The tank initially contains 60 kg of salt.
(b) Amount of salt after 4.5 h

(i) Set up an expression for the rate of change of salt concentration.

(ii) Integrate the expression

(iii) Find the constant of integration

(iv) Solve for A as a function of time.

(v) Calculate the amount of salt after 4.5 h
a. Convert hours to minutes

b.Calculate the concentration

c. Calculate the volume
The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.
The volume added in 4.5 h is

Total volume in tank = 1000 L + 810 L = 1810 L
d. Calculate the mass of salt in the tank

(c) Concentration at infinite time

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.
The graph below shows how the concentration of salt varies with time.