Question

Please help me with this, I have no idea how to do find out the moles of acid/moles of glucose fermented.

Source	Gas in Durham’s Tube	Benedict’s Test (Color)	Benedict’s Test Interpretation	Titration (mL)titrant	Moles of acid/moles of glucose fermented
Control	-	orange	1-1.5% reducing sugar	13.2mL
Homofermentative Lactobacillus	-	green	0.1-0.5% reducing sugar	75.9mL
Heterofermentative Lactobacillus	+	green	0.1-0.5% reducing sugar	129.1mL

Subtract the volume of 0.1 N NaOH titrant used to titrate the uninoculated control from the volume of 0.1 N NaOH titrant used to titrate each of the inoculated cultures. Assuming that the remaining titratable acid is lactic acid produced via fermentation, calculate the moles of acid produced per mole of glucose fermented and compare the values determined for each culture.

Answer 1

Concept: Moles of acid/moles of glucose fermented = the moles of the titrant (NaOH).

For NaOH solution, normality = molarity because equivalent weight = molecular weight

(i) Moles of acid/moles of glucose fermented = (13.2/1000) L * 0.1 mol/L = 0.00132 mol

(ii) Moles of acid/moles of glucose fermented = (75.9/1000) L * 0.1 mol/L = 0.00759 mol

(ii) Moles of acid/moles of glucose fermented = (129.1/1000) L * 0.1 mol/L = 0.01291 mol