Asked by Kate
A 0.205 g sample of impure NaOH requires 17.5 mL of 0.2180 M HCl for neutralization. What is the percent of NaOH in the sample, by weight?
Answers
Answered by
bobpursley
moles acid=.2180*.0175
mass NaOH=formulamassNaOH*molesacid
mass NaOH=40*moles acid above
percent= massNaOH/.205
mass NaOH=formulamassNaOH*molesacid
mass NaOH=40*moles acid above
percent= massNaOH/.205
Answered by
Kate
Thank you! That makes much more sense.
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