56.75mL of 0.256 HI M reacts with 10.00mL sample of NaOH, what is the molarity of sodium hydroxide?

1.45M

Explanation:

The definition of molarity is moles/liter.  The neutralization of NaOH with HI is:

HI + NaOH = NaI  + H2O

One mole of HI reacts with 1 mole of NaOH.  We'll assume this is a titration reaction and that the 10.00ml sample of NaOH contains the same number of moles as the 56.75ml of 0.256M HI.

Moles HI:  (0.256 moles/liter)*(0.05675 L) = 0.01453 moles HI

That means we muct have 0.01453 moles NaOH in 10.0ml of NaOH solution.

(0.01453 moles NaOH)/(0.010L) = 1.45 M

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Another approach is to use the relationship M1V1 = M2V2, which is useful for titrations (M is concentration and V is volume):

We want M2, so rearrange:  M2 = M1V1/V2

M2 = (0.256M)*(56.75ml)/(10.0ml)

M2 = 1.45M