A sample of solid monoprotic acid with molar mass equal to 169.7 g/mol was titrated with 0.1599M sodium hydroxide solution. Calculate the mass in grms fo acid to be used if the volume of NAOH to be used is 25ml.

My calculation

0.1599 x 0.025L = 0.00399
0.00399 x 1000 = 3.9975
3.998 x 169.7 = 678.4 changes gm
x= 0.678gm

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  1. Except for the number of significant figures in your answer I agree with your answer but I don't understand exactly what you did.
    First, since 0.1599 x 0.250 = 0.0039975, rounding to 0.00399 loses one significant figure. Multiplying by 1000 gets that back, I see, but 3.9975 actually should be 4. But why multiply by 1000. You have mols and that's what you need. And what is a "changes gm"? That could be milligram but then you rounded grams off to one fewer s.f. than is allowed. The answer is 0.67838 g which rounds to 0.6784 g (and all of this means I assumed the volume as 25.00 mL).

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