What is the percent by mass of glucose in a solution containing 0.15 moles of glucose dissolved in 410g acetic acid? The molar mass of glucose is 180g/mol
0.15 moles glucose = 0.15 x 180 = 27.00 grams.
mass percent = g solute/100 g soln.
[27.00 g glucose/(27.0+410.0)]*100 = ??
Well, that's quite a mix you've got there! Let's break it down step by step.
First, we need to find the mass of glucose in the solution. To do this, we multiply the number of moles of glucose (0.15 moles) by its molar mass (180 g/mol):
Mass of glucose = 0.15 moles x 180 g/mol = 27 g
Next, we need to calculate the total mass of the solution. This is the mass of acetic acid plus the mass of glucose:
Total mass of solution = Mass of acetic acid + Mass of glucose = 410 g + 27 g = 437 g
Finally, we can find the percent by mass of glucose in the solution. We divide the mass of glucose by the total mass of the solution and multiply by 100 to get the percentage:
Percent by mass of glucose = (Mass of glucose / Total mass of solution) x 100
Percent by mass of glucose = (27 g / 437 g) x 100 ≈ 6.18%
So, the percent by mass of glucose in the solution is approximately 6.18%. That's one sweet addition to the acetic acid!
To find the percent by mass of glucose in the solution, we need to calculate the mass of glucose and divide it by the total mass of the solution, then multiply by 100 to get the answer as a percentage.
Given:
- Moles of glucose (n) = 0.15 moles
- Molar mass of glucose (M) = 180 g/mol
- Mass of acetic acid (m) = 410 g
Step 1: Calculate the mass of glucose using the number of moles and the molar mass.
Mass of glucose (m glucose) = n * M
= 0.15 moles * 180 g/mol
= 27 g
Step 2: Calculate the total mass of the solution by adding the mass of glucose and the mass of acetic acid.
Total mass of the solution (m solution) = m glucose + m acetic acid
= 27 g + 410 g
= 437 g
Step 3: Calculate the percent by mass of glucose in the solution.
Percent by mass of glucose = (mass of glucose / total mass of solution) * 100
= (27 g / 437 g) * 100
≈ 6.18%
Therefore, the percent by mass of glucose in the solution is approximately 6.18%.