Question

informations to solve the problem

supooosed 3.109 g of sodium hydroxide was weighed out,dissolved, quantitively transferred to a 250 ml volumetric flask, made up to the mark,and mixed well, what is the molarity of sodium hydroxide? the answer is 0.3109 M

1.a.

Burets are used to transfer titrant in a titration. A burette is depicted in the lab manual. A 10.00 mL sample of vinegar was titrated with the sodium hydroxide sample from question 1. The initial volume of sodium hydroxide titrant in the buret was 2.50 mL, and the volume at the equivalence point was 32.08 mL.

calculate the net volume of titrated delivered ?

Calculate the moles of sodium hydroxide delivered. Report your answer using the appropriate number of significant figures. Do not include units in your answer.

------------------

b.Calculate the moles of acetic acid reacted.  Report your answer using the appropriate number of significant figures. Do not include units in your answer.

c.Calculate the concentration of acetic acid in the 10.00 mL sample  Report your answer using the appropriate number of significant figures. Do not include units in your answer.

d.Calculate the mass of acetic acid (in grams) reacted. Report your answer using the appropriate number of significant figures. Do not include units in your answer.

Answer 1

Mass (g) 1 @). Molarity (M) = moles of solute (Mault) volume of solur in litres Molar mass y vol. in citres (g/mol) Molarity (M) = 3.109g 0.3109 40 g/mol x 0.250L Volume delivered = final reading - Initial reading = 32.08m2 - 250 ml = 29.58 mL = 0.029582 Moles of MaoH delivered = molarity (M) x volume used in litres - 0.3109M X 0.02458C = 0.0092 b). CH₃COOH + NaOH CH₃COONa + H₂O acetic and Imole of Maoh reacts with Imole of acetic aud. Then moles of acetic auid reacted = moles of NGOH used = 0.0092 conc. of Acetic and in loml = . Moles volume in litres 0.0092 0.010 T = 0.92 mass of acetic acid in grams = moles x molar mas = 0.0092 motes X 60g mole 0.552 Please Rate... -- -