Abeverage mixture contains citric acid, c₆h₈o₇ (molar mass 192 g/mol), plus sugar and flavorings. the mass of citric acid in a sample was determined by titration with sodium hydroxide solution. if a 1.500 g sample of the beverage mix required 46.65 ml of 0.1068 n naoh for complete neutralization, how many grams of citric acid were initially present in the sample?

0.319 gram

Explanation:

Convert Normality to Molarity.

for NaOH: 0.1068 N = 0.1068 M

Citric acid contains 3 groups of COOH so that 1 mole of citric acid reacts with 3 moles of NaOH.

#moles of NaOH =  Concentration x Volume = 0.1068 x 46.65 x 10^-3

= 4.98 x 10^-3 mol

#moles of Citric acid = #moles of NaOH/3 = 1.66 x 10^-3 mol

mass of citric acid = mole x molar mass =  1.66 x 10^-3 x 192 = 0.319 gram

100c is waters boiling point

l dont know l am s sorry