Answer:
To prepare 1 gram of 12.5% ointment using 5% and 20% boric acid ointments, you will need approximately 0.8333 grams of the 5% ointment and 1.1667 grams of the 20% ointment.
Step-by-step explanation:
To determine the grams of each ointment strength needed to prepare 1 g of 12.5% ointment, we can set up a simple equation using the concept of the concentration of solutions.
Let's assume the 5% ointment is represented by "x" grams, and the 20% ointment is represented by "y" grams.
The equation can be set up as follows:
(5% of x) + (20% of y) = (12.5% of 1 g)
Converting the percentages to decimal form:
(0.05x) + (0.20y) = (0.125 * 1)
Simplifying:
0.05x + 0.20y = 0.125
Since we have two variables, we need another equation to solve for both "x" and "y". We can use the fact that the total weight of the ointments should equal 1 g:
x + y = 1
Now we have a system of equations:
0.05x + 0.20y = 0.125
x + y = 1
To solve this system, we can use substitution or elimination. Let's use substitution:
x = 1 - y
Substituting this value of x into the first equation:
0.05(1 - y) + 0.20y = 0.125
0.05 - 0.05y + 0.20y = 0.125
Combining like terms:
0.15y - 0.05 = 0.125
0.15y = 0.125 + 0.05
0.15y = 0.175
Dividing by 0.15:
y = 0.175 / 0.15
y = 1.1667
Substituting the value of y back into x = 1 - y:
x = 1 - 1.1667
x = -0.1667
Since we can't have negative quantities, we discard the negative value of x.
Therefore, we need approximately 1.1667 grams of the 20% ointment and 0.8333 grams of the 5% ointment to prepare 1 g of 12.5% ointment.