If a pharmacist mixed 1 pint of propylene glycol having a specific gravity of 1.20 with 500 mL of water, how many milliliters additional propylene glycol should be added to change the specific gravity to 1.15?
Please help with how to do this problem. I've tried many alligation steps and can't seem to get the answer.
Thank you for your help. The answer in the textbook is 1031.38 mL propylene glycol. There is no detailed explanation in the textbook on how to get to the answer. Also, 1 pint = 473 mL.
Well, I'm no pharmacist, but I'll do my best to help you out with this problem!
Let me break it down for you. We have a mixture of propylene glycol and water, and we want to adjust the specific gravity.
To solve this problem, we need to set up an equation based on the knowledge that the specific gravity is equal to the density of the mixture divided by the density of water.
Now, since density is mass per unit volume, we need to consider the mass of the propylene glycol and the mass of the water.
To make it more fun, let's imagine that the propylene glycol and water are having a wrestling match in the pharmacy. The propylene glycol wants to be the champion and have a specific gravity of 1.15, so it needs some reinforcement.
Let's assume that the density of pure propylene glycol is 1.20 g/mL and the density of water is 1.00 g/mL.
The first step is to calculate the total mass of the initial mixture. We have 1 pint of propylene glycol, which is approximately 473 mL. So, the mass of the initial mixture can be calculated as follows:
mass_initial = (volume_propylene glycol * density_propylene glycol) + (volume_water * density_water)
mass_initial = (473 mL * 1.20 g/mL) + (500 mL * 1.00 g/mL)
mass_initial = 567.6 g + 500.0 g
mass_initial = 1067.6 g
Now, let's see how much additional propylene glycol we need to add to achieve a specific gravity of 1.15.
We can set up the equation:
specific gravity_new = (mass_initial + mass_additional propylene glycol) / (volume_total)
Given that specific gravity_new is 1.15 and volume_total is 973 mL (473 mL initial propylene glycol + 500 mL water), we can solve for mass_additional propylene glycol:
1.15 = (1067.6 g + mass_additional propylene glycol) / 973 mL
Now, I know you've been trying the alligation method, but let me clown around a bit and tell you a secret. Alligation isn't really going to help you here. This is just a simple algebraic equation that you need to solve for mass_additional propylene glycol.
Once you solve for mass_additional propylene glycol, you can use the density of propylene glycol to convert it back to mL.
So, put on your algebraic thinking cap and solve that equation to find out how many milliliters of additional propylene glycol you need to add.
And remember, in the world of pharmacy wrestling matches, the propylene glycol always gets its way! Good luck!
To solve this problem, we can use the fact that the specific gravity of a solution is directly proportional to the concentration of the solute.
Let's start by defining some variables:
- Let x be the number of milliliters of additional propylene glycol needed.
- Let V1 be the initial volume of propylene glycol (1 pint = 473.2 mL).
- Let V2 be the initial volume of water (500 mL).
We can set up a proportion based on the fact that the specific gravity is directly proportional to the concentration:
(V1 + x) / (V1 + V2 + x) = 1.15 / 1.20
Now we can solve for x.
First, cross multiply the equation:
1.15(V1 + V2 + x) = 1.20(V1 + x)
Next, distribute the values:
1.15V1 + 1.15V2 + 1.15x = 1.20V1 + 1.20x
Now, rearrange the equation to isolate x:
1.20x - 1.15x = 1.20V1 - 1.15V1 + 1.15V2
Combine like terms:
0.05x = 0.05V1 + 1.15V2
Finally, divide both sides of the equation by 0.05 to solve for x:
x = (0.05V1 + 1.15V2) / 0.05
Now plug in the values for V1 and V2:
x = (0.05 * 473.2 + 1.15 * 500) / 0.05
x = (23.66 + 575) / 0.05
x = 598.66 / 0.05
x = 11,973.2 mL
Therefore, you would need to add approximately 11,973.2 mL of additional propylene glycol to change the specific gravity to 1.15.
To solve this problem, we can use the concept of specific gravity and the method of alligation. Let's break down the steps to find the solution:
Step 1: Understand the problem and gather information.
- The pharmacist mixed 1 pint (16 fluid ounces) of propylene glycol with a specific gravity of 1.20.
- This mixture is combined with 500 mL of water.
- We need to determine how many milliliters of additional propylene glycol should be added to change the specific gravity to 1.15.
Step 2: Calculate the initial specific gravity of the mixture.
To find the initial specific gravity, we need to find the weighted average based on the amount and specific gravity of the ingredients.
The propylene glycol (PG) has a specific gravity of 1.20, and the water has a specific gravity of 1.00 (since it's considered pure water). We can represent these two ingredients as:
PG (1.20) | Water (1.00)
Now, we need to find the weighted average of the specific gravity. The formula for the weighted average can be used:
Weighted Average = (Quantity of PG * Specific Gravity of PG + Quantity of Water * Specific Gravity of Water) / Total Quantity
In this case, we have 16 fluid ounces or approximately 473 mL of PG and 500 mL of water. Plugging these values into the formula, we get:
Weighted Average = (473 * 1.20 + 500 * 1.00) / (473 + 500)
= (567.6 + 500) / (973)
= 1067.6 / 973
≈ 1.097
So, the initial specific gravity of the mixture is approximately 1.097.
Step 3: Calculate the additional milliliters of propylene glycol needed to change the specific gravity to 1.15.
Using the alligation method, we can visually represent the initial specific gravity of 1.097 on a line.
1.097
|
(1.20) PG
Then, we represent the desired specific gravity of 1.15 on the line.
1.097
|
(1.20) PG
|
(1.15)
To find the ratio between the two solutions, we subtract the specific gravity of 1.15 from 1.097. This gives us:
1.097 - 1.15 = -0.053
Next, we create a proportion by setting up the equation:
Quantity of PG / Total Quantity = Ratio of SG Changes
Let's use x to represent the quantity of additional PG needed. Therefore:
x / (500 + x) = -0.053
Cross-multiplying, we get:
x = -0.053 * (500 + x)
Simplifying:
x = -26.5 - 0.053x
1.053x = -26.5
x ≈ 25.17
Therefore, approximately 25.17 milliliters of additional propylene glycol should be added to change the specific gravity of the mixture to 1.15.
Please note that due to the rounding in the intermediate steps, the final answer is also rounded.
you want this basic algebra equation (first change pint to 483ml)
483ml*1.20+V*1.20+500*1=(V+483+500)1.15
solve for V of additional ml of glycol.
V(1.20-1 )=1.15(983)-1.2(483)-500
V=I get 254ml
check that.