A student placed 17.5 g of glucose 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 calculated the initial molar concetration which is 0.9714 M. I don't know what to do next.
To find the number of grams of glucose in 100 mL of the final solution, you need to use the dilution equation:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
We already know that the initial volume is 60.0 mL and the final volume is 0.500 L.
Next, we need to calculate the final concentration C2. To do this, we'll rearrange the dilution equation:
C2 = (C1V1) / V2
Now let's plug in the values we know:
C1 = 0.9714 M
V1 = 60.0 mL = 0.0600 L
V2 = 0.500 L
C2 = (0.9714 M * 0.0600 L) / 0.500 L
C2 = 0.1166 M
The final concentration of the glucose solution is 0.1166 M.
Finally, to find the number of grams of glucose in 100 mL of the final solution, we can use the formula:
C = n / V
Where C is the concentration in moles per liter, n is the number of moles, and V is the volume in liters.
We rearrange the formula to solve for n:
n = C * V
Now let's plug in the values:
C = 0.1166 M
V = 0.100 L (we convert 100 mL to liters by dividing by 1000)
n = 0.1166 M * 0.100 L
n = 0.01166 moles
Finally, to find the number of grams, we use the molar mass of glucose, which is 180.16 g/mol:
grams = moles * molar mass
grams = 0.01166 moles * 180.16 g/mol
grams = 2.11 g
Therefore, there are 2.11 grams of glucose in 100 mL of the final solution.
To solve this problem, we first need to calculate the molarity of the final solution after the dilution. We can then use this molarity to determine the number of grams of glucose in 100 mL of the final solution.
Given:
Initial volume (V1) = 60.0 mL
Final volume (V2) = 0.500 L
To find the molarity (M2) of the final solution, we can use the dilution formula:
M1V1 = M2V2
Substituting the given values:
(0.9714 M)(60.0 mL) = M2(0.500 L)
Now, let's solve for M2:
M2 = (0.9714 M)(60.0 mL) / (0.500 L)
M2 = 116.568 M mL / L
M2 = 116.568 M
So, the molarity of the final solution after dilution is 116.568 M.
Next, to calculate the grams of glucose in 100 mL of the final solution, we can use the equation:
g = M × V × MW
Where:
g = grams of glucose
M = molarity of the solution (in mol/L)
V = volume of the solution (in L)
MW = molecular weight of glucose
Given:
M = 116.568 M
V = 100 mL = 0.100 L
We need to find the molecular weight (MW) of glucose:
C6H12O6
(6 × Atomic weight of C) + (12 × Atomic weight of H) + (6 × Atomic weight of O)
(6 × 12.01 g/mol) + (12 × 1.008 g/mol) + (6 × 16.00 g/mol)
= (72.06 g/mol) + (12.096 g/mol) + (96.00 g/mol)
≈ 180.156 g/mol
Now, let's calculate the grams of glucose:
g = (116.568 M)(0.100 L)(180.156 g/mol)
g ≈ 2098.61 g
Therefore, there are approximately 2098.61 grams of glucose in 100 mL of the final solution.