[needy]

It has been awhile since I did a molality question, if someone could check my Math I would greatly appreciate it.

Titration of 0.824 grams of KHP (204.22g/mol) required 38.314 grams of NaOH solution to reach a phenolphthalein endpoint.  Find the molality of the NaOH solution .

This is what I did.  change KHP to moles = 0.00403 g KHP = 0.00403 g NaOH

Molality = 0.00403 g NaOH/0.03814 Kg = 0.106 m

Based upon the solution  find the concentration of H2SO4 solution in mol/kg if a 10.063 g aliquot of H2SO4 solution required 57.911 grams of NaOH solution to reach the phenolphthalein endpoint.

m x 0.010036 kg = 0.106 moles x 1 kg
moles of H2SO4 = 10.5 moles
m=10.5 moles H2SO4/0.010063 kg + 1 kg = 10.4 m H2SO4

If I did it wrong any hints would be appreciated. 

[Borek]

change KHP to moles = 0.00403 g KHP = 0.00403 g NaOH

g? Or moles?

Molality = 0.00403 g NaOH/0.03814 Kg = 0.106 m

No, 0.038314 is a mass of solution, not solvent.

[needy]

oops you are right it should have been 0.00403 mol NaOH.  But what is the mass of solvent?

[Borek]

Solution consist of solvent and solute.

[needy]

Oh boy I am so out of it, what was I thinking going back to school at my age? Ok last attempt.
0.0043 moles NaOH = 0.161 g NaOH
38.314 grams of solution NaOH - 0.161 g solute NaOH = 38.153 g solvent
so maybe then this is correct
m=0.00403 moles of NaOH/0.038153 kg = 0.106 m NaOH

[savy2020]

That's right
Proceed with the next part.

[needy]

You guys are the best, off to email my class!!