If water is added to a 25.0 mL sample of 2.70 g/L NaOH(aq) until the volume becomes 4.00 L, find the concentration of the final solution in mol/L.

2.70g NaOH = 0.0675 moles

That means your NaOH solution is .0675M
adding water to make it 4L means you have increased the volume by a factor of 4/.025 = 160
So, the concentration is reduced by a factor of 160 to 0.00042M

To find the concentration of the final solution in mol/L, we need to use the dilution formula:

C1V1 = C2V2

Where:
C1 is the initial concentration in mol/L,
V1 is the initial volume in L,
C2 is the final concentration in mol/L, and
V2 is the final volume in L.

In this case, we have:
C1 = 2.70 g/L NaOH(aq) (but we need to convert this to mol/L)
V1 = 25.0 mL (but we need to convert this to L)
C2 = ? (what we're trying to find)
V2 = 4.00 L

To begin, let's convert the initial concentration from g/L to mol/L. The molar mass of NaOH is 22.99 g/mol for Na, 16.00 g/mol for O, and 1.01 g/mol for H. Therefore, the molar mass of NaOH is 22.99 + 16.00 + 1.01 = 40.00 g/mol.

Converting the initial concentration:
2.70 g/L ÷ 40.00 g/mol = 0.0675 mol/L

Next, we need to convert the initial volume from mL to L:
25.0 mL ÷ 1000 mL/L = 0.0250 L

Now we can use the dilution formula to find the final concentration:
C1V1 = C2V2
(0.0675 mol/L)(0.0250 L) = C2(4.00 L)

Solving for C2:
C2 = (0.0675 mol/L)(0.0250 L) / 4.00 L
C2 = 0.000421875 mol/L

Therefore, the concentration of the final solution is approximately 0.000421875 mol/L.

To find the concentration of the final solution in mol/L, we need to determine the number of moles of NaOH present in the 25.0 mL sample and then divide it by the final volume of 4.00 L.

First, let's calculate the number of moles of NaOH in the initial 25.0 mL sample.

To do this, we can use the formula:
moles = mass / molar mass

The molar mass of NaOH is:
Na: atomic mass = 22.99 g/mol
O: atomic mass = 16.00 g/mol
H: atomic mass = 1.01 g/mol

Total molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Now, let's calculate the mass of NaOH in the 25.0 mL sample using the given concentration:
mass = volume * concentration

Converting mL to L:
25.0 mL = 25.0 mL * (1 L / 1000 mL) = 0.025 L

mass = 0.025 L * 2.70 g/L = 0.0675 g

Now we can find the number of moles of NaOH in the initial 25.0 mL sample:
moles = 0.0675 g / 40.00 g/mol = 0.00169 mol

Finally, we can calculate the concentration of the final solution by dividing the number of moles by the final volume:
concentration = moles / volume

concentration = 0.00169 mol / 4.00 L = 0.0004225 mol/L

Therefore, the concentration of the final solution is approximately 0.0004225 mol/L.