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If 200. mL of water is added to 10.0 mL of 4.00 M NaCl solution,

is the concentration of the dilute solution?

1 Answer

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Answer:

Therefore, the concentration of the resulting dilute solution is 0.190 M.

Step-by-step explanation:

To determine the concentration of the resulting dilute solution, we can use the equation:

M1V1 = M2V2

where M1 and V1 are the initial concentration and volume of the solution, and M2 and V2 are the final concentration and volume of the solution.

In this case, the initial solution is 10.0 mL of 4.00 M NaCl, which can be expressed as:

M1 = 4.00 M

V1 = 10.0 mL = 0.0100 L

The final volume of the solution is the sum of the initial volume and the volume of water added:

V2 = V1 + V_water

V2 = 0.0100 L + 0.200 L

V2 = 0.2100 L

Since the amount of NaCl in the solution does not change upon dilution, we can set up the equation and solve for the final concentration:

M1V1 = M2V2

(4.00 M)(0.0100 L) = M2(0.2100 L)

M2 = (4.00 M)(0.0100 L)/(0.2100 L)

M2 = 0.190 M

Therefore, the concentration of the resulting dilute solution is 0.190 M.

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