If 200 mL of a 2.50 M #NaOH# solution is diluted to 500 mL, what is the new concentration of NaOH?

Answer 1

The new concentration is #1.00*mol*L^-1#.

#"Concentration"# #=# #"Moles of solute"/"Volume of solution"#
We use this relationship to get the initial number of moles, by taking the product,#"concentration"xx"volume"#, and then dividing this molar quantity by the new volume, which we get by simply adding the 2 given volumes.
And thus, #(200xx10^-3cancelLxx2.50*mol*L^-1)/(500xx10^-3cancelL)#
And we get an answer with units #mol*L^-1# as we require.
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Answer 2

The new concentration of NaOH after diluting 200 mL of a 2.50 M solution to 500 mL can be calculated using the formula for dilution:

[C_1V_1 = C_2V_2]

Where: (C_1) = initial concentration (V_1) = initial volume (C_2) = final concentration (V_2) = final volume

Substituting the given values:

(C_1 = 2.50 , \text{M}) (V_1 = 200 , \text{mL}) (V_2 = 500 , \text{mL})

[2.50 , \text{M} \times 200 , \text{mL} = C_2 \times 500 , \text{mL}]

[500 = C_2 \times 500]

[C_2 = \frac{500}{500}]

[C_2 = 1.00 , \text{M}]

So, the new concentration of NaOH after dilution is 1.00 M.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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