How many mL of water would you add to 32 mL of 0.62 M ammonium chloride stock solution to make a 0.3 M solution?
0.62 M x (32/x) = 0.3 M
x = total voume so x-32 will be how much you must add.
I would have .62 mols in a liter
so I have .62 (.032 ) = .01984 mols
.01984 mols /x liters = 0.3 mols/1 liter
x = .01984/.3 = .06613 liters = 66.13 mL
so add 66 -32 = 34 mL of water
How many mL of water would you add to 32 mL of 0.62 M ammonium chloride stock solution to make a 0.3 M solution? >>
Another method. You are diluting it .62/.3 = 2.06 times. That means one part original, and then add 1.06 parts water.
if one part is 32ml, then 1.06 parts is 33.9 ml of water
To find out how much water you need to add to the ammonium chloride stock solution, you can use the formula for dilution:
C1V1 = C2V2
where:
C1 = concentration of the initial solution (0.62 M ammonium chloride)
V1 = volume of the initial solution (32 mL)
C2 = desired concentration (0.3 M ammonium chloride)
V2 = final volume of the diluted solution (V1 + V_water)
First, let's rearrange the formula to solve for V_water:
V_water = V2 - V1
Now, substitute the given values into the equation:
C1V1 = C2V2
(0.62 M)(32 mL) = (0.3 M)(V2)
Next, rearrange the equation to solve for V2:
V2 = (C1V1)/(C2)
V2 = (0.62 M)(32 mL) / (0.3 M)
Calculate V2:
V2 = (0.62 M)(32 mL) / (0.3 M)
V2 = (19.84 mL) / (0.3)
V2 ≈ 66 mL
So, to make a 0.3 M ammonium chloride solution, you would need to add approximately 66 mL of water to 32 mL of the 0.62 M stock solution.