a 69.3g sample of oxalic ac acid, H2C2O4, was dissolved in 1.000L of solution. How would yo prepare 1.00L of 0.150M H2C2O4 from this solution?
figure moles in 69.3g.
moles: 69.3/90=you do it.
Molarity of original solution: 69.3/90=.77M
So you want to decrease molarity to .150, you do that by diluting with water.
You want to dilute it .77/.150 times, or 5.13 times, which means 4.13 parts water, and 1 part original solution.
making one liter, you need each part to be 1000ml/5.13=195ml
Take 195 ml of the original solution, and add water to the 1 liter mark, stir, label, and properly store.
check my figuring.
Well, here's a fun little chemistry puzzle for you! To prepare 1.00L of a 0.150M H2C2O4 solution from your 69.3g sample, we'll need to do a little math and a pinch of chemistry magic.
First, we need to find out how many moles of H2C2O4 are in the 69.3g sample. The molar mass of H2C2O4 is approximately 90.04g/mol, so dividing 69.3g by the molar mass gives us about 0.77 moles of H2C2O4.
Next, we'll use the formula for molarity (M = moles/volume in liters) to figure out the volume that contains 0.77 moles at a concentration of 0.150M. Rearranging the formula, we get volume = moles/M. Plugging in the values, volume = 0.77 moles / 0.150M = about 5.13L.
Now comes the magic part. Since we only need 1.00L of the 0.150M H2C2O4 solution, we can simply dilute the 5.13L of the initial solution to the desired volume by adding the appropriate amount of solvent (water). Just make sure the total volume is exactly 1.00L, and voila! You've prepared your 0.150M H2C2O4 solution.
Always remember, chemistry can be puzzling, but just like a good joke, with a bit of calculation and a dash of creativity, you'll find your way!
To prepare 1.00L of 0.150M H2C2O4 solution from the given solution, you would need to perform a dilution.
Here are the steps you can follow to prepare the desired solution:
1. Determine the number of moles of H2C2O4 in the original solution.
- The molar mass of H2C2O4 is 90.03 g/mol.
- Convert the given mass of the sample to moles.
moles = mass / molar mass
moles = 69.3 g / 90.03 g/mol
2. Calculate the volume of the original solution containing the moles of H2C2O4.
- Use the given concentration of the original solution to find the volume.
volume = moles / concentration
volume = moles / (1.000 L)
3. Calculate the volume of the original solution needed to prepare the desired solution.
- Use the dilution formula:
C1V1 = C2V2
Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Rearrange the formula to solve for V1:
V1 = (C2 * V2) / C1
4. Substitute the values into the equation to find V1.
- C1 = concentration of the original solution (given as 1.000M)
- C2 = concentration of the desired solution (0.150M)
- V2 = final volume of the desired solution (1.00L)
V1 = (0.150 M * 1.000 L) / (1.000 M)
5. Calculate the volume of the solvent (water) needed to make up the difference in volumes between the original and desired solutions.
- Subtract the volume of the original solution from the volume of the solvent needed.
volume of solvent = V2 - V1
6. Add the calculated volume of the original solution to the calculated volume of the solvent to get the total volume of the desired solution.
Follow these steps, and you will be able to prepare 1.00L of 0.150M H2C2O4 solution from the given solution.
To prepare 1.00L of a 0.150M H2C2O4 solution from the given solution, we need to calculate the volume of the stock solution required and then dilute it to the desired concentration. Here's how you can do it:
1. Determine the number of moles of H2C2O4 in the stock solution:
- The molar mass of oxalic acid (H2C2O4) is 90.03 g/mol.
- The sample has a mass of 69.3 g.
- Divide the mass by the molar mass to find the number of moles:
Moles = mass / molar mass = 69.3 g / 90.03 g/mol ≈ 0.770 mol
2. Calculate the volume of the stock solution required using the formula:
Moles = Molarity × Volume (in liters)
Rearrange the formula to solve for volume:
Volume (in liters) = Moles / Molarity
Volume = 0.770 mol / 0.150 M ≈ 5.133 L
Therefore, you need 5.133 liters of the stock solution to prepare 1.00L of 0.150M H2C2O4.
3. To make the desired 1.00L of 0.150M H2C2O4 solution, pour the required volume of the stock solution (5.133L) into a container and then carefully add distilled water until the total volume reaches 1.00L. Mix well to ensure homogeneity.
Now, you have prepared 1.00L of a 0.150M H2C2O4 solution from the given stock solution.