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.