0.500 WHAT? liters?
11.5 g/100 = 0.115 g/mL original.
25 mL x that = grams in 500 mL.
Divide by 5 to find grams in 100 mL.
11.5 g/100 = 0.115 g/mL original.
25 mL x that = grams in 500 mL.
Divide by 5 to find grams in 100 mL.
C1V1 = C2V2
where:
C1 is the initial concentration of the glucose solution,
V1 is the initial volume of the glucose solution,
C2 is the final concentration of the diluted solution,
V2 is the final volume of the diluted solution.
Let's calculate it step by step.
Step 1: Determine the initial volume and concentration of the glucose solution.
Initial volume (V1) = 11.5 mL
Initial concentration (C1) = ?
Step 2: Determine the final volume of the diluted solution.
Final volume (V2) = 100 mL
Step 3: Determine the final concentration of the diluted solution.
Using the formula:
C1V1 = C2V2
Rearranging the formula:
C2 = (C1V1) / V2
Substituting the known values:
C2 = (11.5 mL * C1) / 100 mL
Step 4: Determine the mass of glucose in the final solution.
To do this, we need to consider the volume of the diluted solution used for the analysis and the final concentration.
Volume used (V3) = 25.0 mL
Final concentration (C3) = 0.500 g/mL
Using the formula:
Mass of glucose = Volume used * Concentration
Mass of glucose = 25.0 mL * 0.500 g/mL
So, the calculation for the number of grams of glucose in 100 mL of the final solution is as follows:
C2 = (11.5 mL * C1) / 100 mL
Mass of glucose = 25.0 mL * 0.500 g/mL
Please provide the value for the initial concentration (C1) of the glucose solution so we can proceed with the final calculation.
Step 1: Determine the concentration of the 25.0 mL solution.
We know that the 25.0 mL solution was diluted to 0.500 M (Molar). This means that there are 0.500 moles of glucose in 1 liter (1000 mL) of the solution.
Step 2: Calculate the number of moles of glucose in the 25.0 mL solution.
Since we know the concentration (0.500 M) and the volume (25.0 mL) of the solution, we can use the formula: Moles = Concentration x Volume (in liters). Converting 25.0 mL to liters, we have 0.0250 L. Therefore, the number of moles of glucose in the 25.0 mL solution is 0.500 M x 0.0250 L = 0.0125 moles.
Step 3: Calculate the number of moles of glucose in 100 mL of the final solution.
Since we diluted the 25.0 mL solution to 100 mL, the concentration remains the same. Therefore, we can use the same number of moles of glucose (0.0125 moles) for the 100 mL solution.
Step 4: Calculate the mass of glucose in 100 mL of the final solution.
To find the mass of the glucose, we need to multiply the number of moles by its molar mass. The molar mass of glucose (C₆H₁₂O₆) is approximately 180 grams/mol. Therefore, the mass of glucose in 100 mL of the solution is 0.0125 moles x 180 g/mol = 2.25 grams.
So, there are 2.25 grams of glucose in 100 mL of the final solution.