IF 121ml of a 1.0 M glucose solution is diluted to 500.0 mL, what is the molarity of the diluted solution?
1.0 M x (121/500) = ?
or you can use c1v1 = c2v2 where
c = concn
v = volume
1.0*121 = ?M*500
Well, if we're talking about dilutions, let's dilute the seriousness of this question with a dash of humor!
Now, to find the molarity of the diluted solution, we can use the formula: M1V1 = M2V2.
So, plugging in the values, we have M1 = 1.0 M, V1 = 121 mL, V2 = 500 mL, and we need to find M2 (the molarity of the diluted solution).
Now, let's do some math...but don't worry, I'll keep it light!
(1.0 M) x (121 mL) = (M2) x (500 mL)
Let's cancel out the mLs and play with the numbers:
M2 = (1.0 M) x (121 mL) / (500 mL)
M2 = (1.0 M) x (0.242)
M2 ≈ 0.24 M (rounded to two significant figures)
So, the molarity of the diluted solution is approximately 0.24 M.
Hope that puts a smile on your face!
To determine the molarity of the diluted solution, you need to use the formula:
M₁V₁ = M₂V₂
Where:
M₁ = initial molarity
V₁ = initial volume
M₂ = final molarity
V₂ = final volume
In this case, the initial molarity (M₁) is 1.0 M, the initial volume (V₁) is 121 ml, the final volume (V₂) is 500.0 ml, and you want to find the final molarity (M₂).
Now let's substitute the values into the formula:
(1.0 M)(121 ml) = (M₂)(500.0 ml)
Rearrange the formula to solve for M₂:
M₂ = (1.0 M)(121 ml) / (500.0 ml)
M₂ = 121 ml / 500.0 ml
M₂ = 0.242 M
Therefore, the molarity of the diluted solution is 0.242 M.