A student placed 17.0 g of glucose (C_6H_{12}O_6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100 mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 55mL sample of this glucose solution was diluted to 0.500L. How many grams of glucose are in 100 mL of the final solution?
Express your answer numerically in grams.
17.0 g in 100 mL.
55 mL taken diluted to 500 mL.
So the glucose placed in the 500 mL flask is 17.0 x (55/100) = ??
Therefore, the concn (grams/mL) in the 500 mL flask is ??/500 mL. Multiply by 100 if you want the grams/100 mL.
Well, well, well, looks like we've got ourselves a chemistry question, huh? Don't worry, I'm here to help you out with a side of humor!
So, let's break it down, shall we? We started with a 17.0 g of glucose in a 100 mL solution. Then, we diluted that solution by adding more water. And finally, we took a 55 mL sample of that diluted solution and diluted it even more to reach a volume of 0.500L. Phew!
Now, the question asks how many grams of glucose are in 100 mL of the final solution. Since we started with 17.0 g in 100 mL and diluted it further, we'll need to do a bit of math (don't worry, it won't be too terrifying).
So, let's calculate it step by step:
(17.0 g / 100 mL) = (x g / 55 mL)
Cross multiply that equation and we get:
(100 mL * x g) = (17.0 g * 55 mL)
Now let's solve for x:
x = (17.0 g * 55 mL) / 100 mL
x = 9.35 g
Ah, so there you have it! In 100 mL of the final solution, there are approximately 9.35 grams of glucose. Hope that helps, and remember, chemistry can be sweet too!
To solve this problem, we need to first determine the concentration of the glucose solution. We can do this by calculating the moles of glucose in the 55 mL sample and then dividing by the total volume of the final solution.
1. Calculate the moles of glucose in the 55 mL sample:
- The molar mass of glucose (C6H12O6) is 180.16 g/mol.
- Convert the given mass of glucose to moles:
moles = mass / molar mass
moles = 17.0 g / 180.16 g/mol
moles = 0.0944 mol
2. Determine the concentration of glucose in the 55 mL sample:
- Concentration (Molarity) = moles / volume
- Convert the volume from mL to L:
volume = 55 mL / 1000 mL/L
volume = 0.055 L
- Concentration (Molarity) = 0.0944 mol / 0.055 L
- Concentration (Molarity) = 1.715 M
3. Use the concentration to calculate the moles of glucose in 100 mL of the final solution:
- Concentration (Molarity) = moles / volume
- Convert the volume from mL to L:
volume = 100 mL / 1000 mL/L
volume = 0.1 L
- moles = Concentration (Molarity) * volume
moles = 1.715 M * 0.1 L
moles = 0.1715 mol
4. Convert the moles back to grams of glucose:
- Grams = moles * molar mass
- Grams = 0.1715 mol * 180.16 g/mol
- Grams = 30.92 g
Therefore, there are 30.92 grams of glucose in 100 mL of the final solution.
To find the number of grams of glucose in 100 mL of the final solution, we can use the concept of dilution.
First, let's calculate the concentration of glucose in the original solution. We know that 17.0 g of glucose was dissolved in water to make a 100 mL solution. Thus, the concentration of glucose in the original solution is:
Concentration (C1) = mass of solute (glucose) / volume of solution
C1 = 17.0 g / 100 mL
Next, we dilute a 55 mL sample of the original solution to a total volume of 0.500L (500 mL). The concentration of the diluted solution is given as:
Concentration (C2) = C1 * V1 / V2
where C2 is the concentration of the diluted solution, V1 is the volume of the original solution, and V2 is the total volume of the diluted solution.
For the dilution, we have:
C2 = C1 * (V1 / V2)
C2 = (17.0 g / 100 mL) * (55 mL / 500 mL)
C2 = (17.0 g * 55) / (100 * 500)
C2 ≈ 1.87 g / 500 mL
To find the number of grams of glucose in 100 mL of the final solution, we can use the proportion:
C2 = x g / 100 mL
Solving for x (the number of grams):
x = C2 * 100 mL
x ≈ (1.87 g / 500 mL) * 100 mL
x ≈ 0.374 g
Therefore, there are approximately 0.374 grams of glucose in 100 mL of the final solution.