A student placed 10.5 g of glucose C6H12O6 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 60.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100 mL of the final solution?

Question

A student placed 10.5 g of glucose C6H12O6 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 60.0 mL sample of this glucose solution was diluted to 0.500 L. How many grams of glucose are in 100 mL of the final solution?

I tried using the formula M1V1=M2V2 but the 60.0 mL is throwing me off? how do i do this problem? thanks!

Answer 1

10.5 g is in a 100; therefore, 10.5/100 = 0.105 g/mL.

Now take 60.0 of that solution is diluted to 500 mL.
So we have 10.5 g/mL x 60 mL = xx grams and that divided by 500 = yy g/mL. Now multiply that by 100 to know how much is in 100 mL. Check my thinking.

Answer 2

To solve this problem, you can use the equation M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.

Let's break it down step-by-step:

Step 1: Calculate the initial molarity (M1).
Given that the student placed 10.5 g of glucose in a volumetric flask with a volume of 100 mL, you can calculate the initial molarity (M1) as follows:

M1 = (mass of glucose) / (molar mass of glucose * volume of solution)
= 10.5 g / (180.16 g/mol * 0.100 L)
= 5.83 M

Step 2: Calculate the final molarity (M2).
Since you diluted a 60.0 mL sample to 0.500 L, the final molarity (M2) can be found using the formula:

M2 = (M1 * V1) / V2
= (5.83 M * 0.060 L) / 0.500 L
= 0.7032 M

Step 3: Calculate the grams of glucose in the 100 mL final solution.
Now that you have the final molarity, you can use it to find the grams of glucose in 100 mL of the final solution. Use the formula:

mass of glucose = molar mass of glucose * (M2 * volume of solution)
= 180.16 g/mol * (0.7032 M * 0.100 L)
= 12.66 g

Therefore, there are 12.66 grams of glucose in 100 mL of the final solution.

Answer 3

To solve this problem, you can use the formula for dilution,

M1V1 = M2V2

where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.

In this case, the initial volume is 10.5 g of glucose in the volumetric flask with enough water to reach the 100 mL mark. The final volume is 100 mL.

First, we need to calculate the initial concentration (M1). To do this, we need to know the molar mass of glucose, which is 180 g/mol.

To find the initial concentration (M1), divide the mass of glucose (10.5 g) by the molar mass (180 g/mol) and divide it by the initial volume (100 mL) converted to liters:

M1 = (10.5 g / 180 g/mol) / (100 mL / 1000 mL/L)
M1 = 0.0583 mol/L

Now, we can use the dilution formula to find the final concentration (M2).

M1V1 = M2V2

(0.0583 mol/L)(60.0 mL) = M2(0.500 L)

M2 = (0.0583 mol/L)(60.0 mL) / (0.500 L)
M2 = 0.6996 mol/L

So, the final concentration of the glucose solution after dilution is 0.6996 mol/L.

Now, to find the grams of glucose in 100 mL of the final solution, we can use the final concentration (M2) and the final volume (100 mL).

First, convert the final volume to liters:

V2 = 100 mL / 1000 mL/L
V2 = 0.100 L

Then, use the formula for moles:

moles = concentration × volume

moles of glucose = (0.6996 mol/L) × (0.100 L)
moles of glucose = 0.06996 mol

Finally, we can calculate the grams of glucose:

grams of glucose = moles × molar mass

grams of glucose = 0.06996 mol × 180 g/mol
grams of glucose = 12.5916 g

Therefore, there are approximately 12.6 grams of glucose in 100 mL of the final solution.