A 3.0 -L sample of a 5.8 M NaCl solution is diluted to 65 L. What is the molarity of the diluted solution?

Oh, I see we're diluting some saltwater. Well, let me grab my watering can of knowledge and give you an answer. So, if we take a 3.0 L sample of a 5.8 M NaCl solution and dilute it to 65 L, we can determine the molarity of the diluted solution.

To calculate the molarity, we need to consider the amount of solute and the volume of the solution. In this case, the initial sample had a volume of 3.0 L and a molarity of 5.8 M. After dilution, we're left with a total volume of 65 L.

To find the molarity of the diluted solution, we can use the formula:

Molarity1 x Volume1 = Molarity2 x Volume2

Plugging in our values, we get:

(5.8 M) x (3.0 L) = Molarity2 x (65 L)

Molarity2 = (5.8 M x 3.0 L) / 65 L

Molarity2 ≈ 0.267 M

So, the molarity of the diluted solution is approximately 0.267 M. Diluting saltwater never tasted so mathematical!

To find the molarity of the diluted solution, you need to use the equation:

M1V1 = M2V2

where:
M1 = initial molarity of the solution
V1 = initial volume of the solution
M2 = final molarity of the solution
V2 = final volume of the solution

In this case, we are given:
M1 = 5.8 M (initial molarity)
V1 = 3.0 L (initial volume)
M2 = ? (final molarity)
V2 = 65 L (final volume)

Plugging in these values into the equation, we get:

(5.8 M)(3.0 L) = M2(65 L)

Simplifying the equation:

17.4 = M2(65)

Now, we can solve for M2 by dividing both sides of the equation by 65:

M2 = 17.4 / 65

Calculating this, we find:

M2 ≈ 0.267 M

Therefore, the molarity of the diluted solution is approximately 0.267 M.