Question

As the salt KNO₃ dissolves in methanol, the number x(t) of grams of the salt in the solution after t seconds satisfies the differential equation: dx/dt = 0.8 times -0.004 times 2 If x = 50 when t = 0. Express x as a function of time t. How long will it take an additional 50 grams of salt to dissolve?

Answer 1

The mass of KNO3 in solution as a function of time is x(t) = -0.0064t + 50. However, under these conditions, an additional 50 grams cannot dissolve, as indicated by a negative dissolution time when solving for when x(t) would equal 100 grams.

Step-by-step explanation:

To solve for x as a function of time t, one must integrate the given differential equation dx/dt = 0.8 × -0.004 × 2, simplifying to dx/dt = -0.0064. Since the rate of dissolution is constant, we can integrate directly:

∫ dx = ∫ -0.0064 dt

x(t) = -0.0064t + C

To find C, we substitute the initial condition x(0) = 50:

50 = -0.0064(0) + C

C = 50

Thus, the function describing the mass of KNO₃ in solution over time is x(t) = -0.0064t + 50.

To find the time it takes for an additional 50 grams to dissolve, set x(t) = 100:

100 = -0.0064t + 50

t = (100 - 50) / -0.0064

t = -781.25 seconds

The negative value indicates that the additional 50 grams of salt cannot dissolve under the given conditions, and thus the solution appears to be already saturated with the initial 50 grams.