in this experiment you are you will begin by preparing a dilute soultion of sodium hydroxide by taking 50 ml of 3 M NaOh stock solution and diluting it with 250 ml of water.

i found that and the answer was .5
but how do i do this

b. calculate the number of moles and grams of oxalic acid reqiored to neutralize 35 ml of the dilute NAOh solution you have made abobe.

First I want to point out a serious error although you may have copied the problem exactly as you received it. But just in case you rephrased it, this will critique that. You do not know the concn of the dilute NaOH BECAUSE you must assume the volumes are additive (the problem states you took 50 mL of the stock solution and diluted it WITH 250 mL of water. This is not the way dilutions are performed. You take 50 mL and add ENOUGH water to make the final volume 250 mL. In the latter case, the concn in the diluted solution is 3 M x (50/250) = 0.60 M. In the former case, you can't calculate the concn unless you know the density of the solution and the final volume.

For the second part, write and balance the equation. I will call oxalic acid, to make things simple, H2Ox
H2Ox + 2NaOH ==> Na2Ox + HOH (water)

How many moles do we have in 35 mL of the dilute NaOH? That is M x L = 0.60 x 0.035 = ??
Notice from the equation that it takes only 1/2 the moles of HOx to neutralize the NaOH; therefore, take the ?? moles NaOH and divide by 2. Thaat will give you the number of moles of HOx.
Then grams H2Ox = moles x molar mass.

To calculate the number of moles and grams of oxalic acid required to neutralize the dilute NaOH solution, you need to follow these steps:

Step 1: Determine the concentration of the diluted NaOH solution.
In the given information, it states that you prepared a dilute solution of sodium hydroxide by taking 50 ml of 3 M NaOH stock solution and diluting it with 250 ml of water. To find the final concentration of the diluted solution, you can use the formula:

C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

In this case, C1 = 3 M (concentration of the stock solution), V1 = 50 ml (volume of the stock solution), C2 = unknown (final concentration of the diluted solution), and V2 = 50 ml (volume of the diluted solution). Substituting these values into the formula gives you:

(3 M)(50 ml) = C2(300 ml)

Solving for C2 gives you:

C2 = (3 M)(50 ml) / (300 ml) = 0.5 M

So, the final concentration of the diluted NaOH solution is 0.5 M.

Step 2: Calculate the moles of NaOH in the diluted solution.
To find the number of moles (n) of NaOH in the diluted solution, you can use the formula:

n = C x V
where C is the concentration and V is the volume.

In this case, C = 0.5 M (concentration of the diluted solution) and V = 35 ml (volume of the diluted solution). Substituting these values into the formula gives you:

n = (0.5 M)(35 ml)

Solving for n gives you:

n = 17.5 mmol (millimoles)

Step 3: Determine the balanced chemical equation for the neutralization reaction between NaOH and oxalic acid.
The balanced chemical equation for the reaction between NaOH and oxalic acid (H2C2O4) is:

2 NaOH + H2C2O4 ⟶ Na2C2O4 + 2 H2O

Step 4: Calculate the moles and grams of oxalic acid required.
From the balanced chemical equation, you can see that the stoichiometric ratio between NaOH and H2C2O4 is 2:1. That means for every 2 moles of NaOH, you need 1 mole of H2C2O4.

Since you found the moles of NaOH in the diluted solution to be 17.5 mmol, the moles of H2C2O4 required will be half that:

moles of H2C2O4 = 17.5 mmol / 2 = 8.75 mmol

To convert moles to grams, you'll need to know the molar mass of oxalic acid (H2C2O4), which is 90.03 g/mol.

grams of H2C2O4 = moles of H2C2O4 × molar mass of H2C2O4
= 8.75 mmol × 90.03 g/mol

Finally, calculate the grams of oxalic acid required.

grams of H2C2O4 = 787.58 mg

Therefore, you would need approximately 787.58 mg (or 0.78758 g) of oxalic acid to neutralize 35 ml of the diluted NaOH solution.