Chemistry! How many grams of lactic acid were in the sample?

moles NaOH = 0.01725 L x 0.155 M=0.00267
because the lactic acid is a momoprotic acid
moles lactic acid = 0.00267
mass lactic acid = 0.00267 mol x 90.08 g/mol=0.241 g

CH3CH(OH)COOH(aq)+NaOH(aq)
->CH3CH(OH)COONa(aq)+H2O(l)

No of moles of NaOH
= [NaOH]x(volume of NaOH in L)
= 0.155x17.25x10^-3
= 2.67375x10^-3

The required ratio of the no of moles of CH3CH(OH)COOH to that of NaOH is 1:1.
Therefore, the no of moles of CH3CH(OH)COOH required
= no of moles of NaOH
= 2.67375x10^-3

Relative Molecular Mass(RMM) of CH3CH(OH)COOH= 90
1 mole of CH3CH(OH)COOH= 90g
2.67375x10^-3 moles of CH3CH(OH)COOH
= 90x2.67375x10^-3
= 0.241g(mass of lactic acid in the sample)

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