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QUESTION

# If 36.10 mL of 0.223 M of NaOH is used to neutralize a 0.515 g sample of citric acid, what is the molar mass of the acid?

The molar mass of the citric acid is "192 g/mol".

The balanced chemical equation for the reaction that takes place between sodium hydroxide, NaOH, and citric acid, C_6H_8O_7, looks like this

C_6H_8O_(7(aq)) + color(red)(3)NaOH_((aq)) -> Na_3C_6H_5O_(7(aq)) + 3H_2O_((l))

Notice the 1:color(red)(3) that exists between citric acid and sodium hydroxide; what this tells you is that, for every mole of citric acid, you need 3 times more moles of sodium hydroxide for the reaction to take place.

Since you know the and the volume of the NaOH you've used, you can calculate how many moles of NaOH reacted

C = n/V => n = C * V

n_(NaOH) = "0.223 M" * 36.10 * 10^(-3)"L" = "0.00805 moles"

Now use the aforementioned mole ratio to see how many moles of citric acid were present in 0.515 g

0.00805cancel("moles NaOH") * "1 mole citric acid"/(3cancel("moles NaOH")) = "0.00268 moles citric acid"

Now simply divide the mass of citric acid given by the number of moles it contained to get the compound's molar mass

M_M = m/n = "0.515 g"/("0.00268 moles") = "192.16 g/mol"

Rounded to three , the number of sig figs given for 0.515 g, the answer will be

M_M = color(green)("192 g/mol")

SIDE NOTE The actual molar mass of citric acid is 192.12 g/mol, so your result is in agreement with the known value.