Describe how you would make 250 mL of 0.500 M HCI from 12.1 M HCI. Show calculations.
Using the dilution formula M1V1 = M2V2 you would use the following calculations.
((250 mL)x(0.500 M HCl))/12.1 M HCl
= 10.3 mL
This means you want to add 10.3 mL of the 12.1 M HCl into a 250-mL volumetric flask and then fill the rest with distilled water up to the etch line in the flask.
How did you get 12.1 M HCI?
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The materials used are as follows:tripple beam balance,evaporating basin,stop watch,HCL pellets,matches,candle, .weigh the empty basin on tripple beam balance and records the mass,then put hcl on evaporating basin,burn it until it change from blue to white,then put it again on tripple beam balance and records the mass,then calculate the mass of empty basin and that with hcl,state the sources of erros after experiments.this is how u can prepare 250 ML in 0.500 M HCL
given 150ml of kcl solution of20g/l.prepare75mlof kcl solution whose concentration has to be 12g/l.
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Describe how you would make 1.50 Liters of 0.555 M magnesium chloride solution from a 3.77 M magnesium chloride solution already prepared?
To make 250 mL of 0.500 M HCl from a stock solution of 12.1 M HCl, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = Concentration of the stock solution
V1 = Volume of the stock solution
C2 = Final concentration
V2 = Final volume
In this case:
C1 = 12.1 M
V1 = ?
C2 = 0.500 M
V2 = 250 mL
Rearranging the formula, we have:
V1 = (C2 * V2) / C1
Substituting the values, we get:
V1 = (0.500 M * 250 mL) / 12.1 M
Calculating this expression gives us:
V1 ≈ 10.33 mL
Therefore, to make 250 mL of 0.500 M HCl from 12.1 M HCl, you will need to mix approximately 10.33 mL of the 12.1 M HCl stock solution with enough solvent (usually water) to reach a total volume of 250 mL.
My answer is 80 mL HCl.
I used the dilution formula:
250 mL x .500 M HCl = "X" mL x 12.1 M HCl
= 80 mL