how would you prepare 250 ml of 0.69m al2(so4)3 solution from a 2.0 al2(so4)3 stock solution
Well, to prepare 250 ml of 0.69M Al2(SO4)3 solution from a 2.0M Al2(SO4)3 stock solution, you would need a bit of dilution magic!
First, we need to find out the ratio of the stock solution and the desired solution.
The formula for dilution is C1 x V1 = C2 x V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the desired concentration, and V2 is the desired volume.
So, let's plug in the numbers. C1 is 2.0M, V1 is ? (we don't know yet), C2 is 0.69M, and V2 is 250 ml.
Now let's solve the equation for V1:
2.0M x V1 = 0.69M x 250 ml
V1 = (0.69M x 250 ml) / 2.0M
V1 = 86.25 ml
So, you need to take 86.25 ml of the 2.0M Al2(SO4)3 stock solution and dilute it with enough solvent (probably water) to reach the final volume of 250 ml.
Remember, safety first, don't juggle with chemicals, and always wear your lab coat!
To prepare 250 ml of a 0.69M Al2(SO4)3 solution from a 2.0M Al2(SO4)3 stock solution, you can follow these steps:
1. Determine the volume of the stock solution needed: The equation C1V1 = C2V2 can be used, where C1 is the concentration of the stock solution, V1 is the volume of the stock solution needed, C2 is the desired concentration, and V2 is the final volume of the solution. Rearranging the equation, we have V1 = (C2 × V2) / C1.
V1 = (0.69M × 250ml) / 2.0M = 86.25ml
Therefore, you will need 86.25ml of the stock solution.
2. Measure 86.25ml of the 2.0M Al2(SO4)3 stock solution using a graduated cylinder or pipette. Be sure to use appropriate safety precautions, such as wearing gloves and goggles.
3. Transfer the measured volume of the stock solution into a 250ml volumetric flask. This flask will allow you to accurately measure the final volume of the solution.
4. Add distilled water to the volumetric flask until the volume reaches the 250ml mark. This will dilute the stock solution and create a 0.69M Al2(SO4)3 solution.
5. Carefully mix the contents of the flask to ensure that the solution is well-mixed.
Now, you have successfully prepared 250ml of a 0.69M Al2(SO4)3 solution from a 2.0M Al2(SO4)3 stock solution.
To prepare 250 ml of a 0.69 M Al2(SO4)3 solution from a 2.0 M Al2(SO4)3 stock solution, we can use the formula:
C1V1 = C2V2
where:
C1 = concentration of stock solution
V1 = volume of stock solution needed
C2 = desired concentration of the final solution
V2 = final volume of the solution
In this case, we can plug in the given values into the formula to solve for V1 (volume of stock solution needed):
(2.0 M) * V1 = (0.69 M) * 250 ml
V1 = (0.69 M * 250 ml) / 2.0 M
V1 ≈ 86.25 ml
Therefore, to prepare 250 ml of a 0.69 M Al2(SO4)3 solution from a 2.0 M Al2(SO4)3 stock solution, you would need to measure approximately 86.25 ml of the stock solution and then dilute it with enough solvent (such as water) to reach a total volume of 250 ml.
2.0 what? m
Is that 0.69m or 0.69M? I assume 0.69m
That means 0.69 mol/0.250 kg solvent.
0.69 x molar mass Al2(SO4)3 = approx 60g.
Then the solvent has a mass of 250 g for a total mass of approx 250+60 = approx 0310 g. What's the density of the solution? volume = mass/density = ? = approx 310 mL or 0.310 L.
Then use c1v1 = c2v2.