Question

1.) It takes 2.085x10^1 mL of 1.30x10^-1 M NaOH to titrate 2.9660x10^-1 grams of solid monoprotic acid (HX). What is the molar mass of this acid. Express in scientific not. w/ 3 sig figs.

Here is my work, but I'm not sure if it's right. Please check. Thanks.
(1.30x10^-1 M NaOH * 2.085x10^1 mL )/1000
=0.0027105
2.9660x10^-1 grams/0.0027105= molar mass= 109.426=1.09x10^2


2.) 2.72x10^-1 grams of solid monoprotic acid (HX) w/ mm of 1.0572x10^2 g/mol was titrated w/ 1.220x10^-1 M NaOH. How many mL of NaOH would it take to reach the endpoint?
Work:
39.997 mm of NaOH
2.72x10^-1 grams / 39.997g NaOH= 0.00680051 moles/1.0572x10^2 g/mol * 1000 ml = 6.43x10^-8

are these correct?

Answer 1

So I double checked, the first is right. Although I'm not sure about 2.)

Answer 2

1 is right.
2 is wrong. In 2 you divided grams HX by molar mass NaOH to find mols

Answer 3

I say 1 is right. I just checked the procedure and not the math.