Dilution and moles

I have two problems that I have trouble with. One I attempted, but I don't think I came up with the right formula. Any help would be appreciated!

Problem one

Use the dilution relationship (Mi x Vi = Mf x Vf) to calculate the volume of 0.500 M NaOH needed to prepare 500 mL of .250 M NaOH.

This is my work

? L Solution = 500mL x (1L/1000mL) = .500L

? mol NaOH = .500 L Solution x (0.250 M NaOH/1 L Solution) = .125 mol NaOH

? L solution = .125 mol NaOH x (1L Solution/.500 mol NaOH) = .25

The problem then asks me to round my answer to the nearest ones place, so i have a feeling this equation has an error. Can anyone guide me to getting it right? Thanks!



Problem 2
49.22 ml of a 2.01 M naOH solution reacts completely with 40.28 mL of HCl solution according to the blanaced chemical reaction shown below:

HCl (aq) + NaOH(aq) -> NaCl(aq) + H2O(I)
 

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Reminder: M means moles per liter, or moles per 1000 milliliters.
Your initial volume is "unknown" and your final volume is expected to be 500 ml.
In symbols as MILLILITERS, Vf=Vi+Vu, where Vu is unknown volume to add.

Simply substitute the values given and use simple algebra, but first keep units in moles per thousand milliliters for M, and milliliters for V.

Start with the concentration formula relationship for ease. Find the unknown initial volume. Use the simple volume addition relationship next.
 
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Problem 2
49.22 ml of a 2.01 M naOH solution reacts completely with 40.28 mL of HCl solution according to the blanaced chemical reaction shown below:

HCl (aq) + NaOH(aq) -> NaCl(aq) + H2O(I)
Heya,

So what is the question? To find the volume of products? Also, do you have any workings out? If not, what do you know about the balanced equations and ratios of reactants that will help?

The Bob
 

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The second question is missing the molarity of the hydrochloric acid. This is what the question then must ask.
 

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