Question

Trial 6 Trial 5 Trial 2 Trial 4 Trial 3 Trial 1 Volume of acid sml solution acquired: smit smi smi 6 мм ISmI 1 omi omi omlomi Initial buret reading: OML Oml Final buret reading: umu 5ml smi u.sms . s u .s. *Volume of NaOH I UML solution used: smi I smi u.5ml 9.5 u.s. I volume trail m NaOH t (trail) 7 mol NaOH MGIHAI > MHCL *Molarity of acid solution: Name: *Average Molarity of acid: Percent Range:

Answer 1

Given , molarity of NaOH solution =$\frac{mass \: of\: NaOH(g)\times 1000}{molar\: mass\: of NaOH(g/mol)\times volume \: of\: solution (mL)}$

= $\frac{2\times 1000}{40\times 250}$ = 0.2 M

now

moles of HCl = moles of NaOH

or,

M_acidV_acid = M_NaOHV_NaOH

so, molarity of acid =  M_NaOHV_NaOH / V_acid = { (molarity of NaOH $\times$ volume of NaOH)/ volume of acid }

	Trial 1	trial 2	Trial 3	Trial 4	Trial 5	Trial 6
molarity of acid solution (M)	(0.2$\times$ 4)/5 = 0.16	(0.2$\times$5)/5 = 0.2	(0.2$\times$5)/5 = 0.2	(0.2$\times$4.5)/5 = 0.18	( 0.2$\times$5.5)/5 = 0.22	( 0.2$\times$4.5)/5 = 0.18

Average molarity of acid = (0.16+ 0.2 +0.2+ 0.18 +0.22+0.18)

= 0.19 M

maximum value = 0.22 M

minimum value = 0.18 M

then range = 0.22 - 0.18 = 0.04

percentage range = $\frac{range \times 100}{average\: molarity}$

= $\frac{0.04\times 100}{0.19}$

= 21.05 %.