Can someone solve this lab for me, please 32 B. Molecular mass of an unknown aci
ID: 548076 • Letter: C
Question
Can someone solve this lab for me, please
Explanation / Answer
Volume of NaOH used = Intital buret - final buret
Volume of NaOH used = V1 = 17 18.6 16.8 ml
= 0.017 0.0186 0.0168 L
No. of moles of NaOHused = V1 * Molarity of NaOH = V1 * 0.1 mol/L
No. of moles of NaOHused = 0.0017 0.00186 0.00168 moles
No. of moles of acid = no. of moles of NaOH used
No. of moles of acid = 0.0017 0.00186 0.00168 moles
GIven mass of unkonwn acid = 0.5 0.5 0.5 g
Molar mass of acid = mass / No. of moles of acid
Molar mass of acid = 294.12 268.82 297.62 g/mol
Average molecular mass of acid = 286.85 g/mol
Deviation = 7.27 - 18.03 10.77
Average deviation = 12.02
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.