What is the normality of a solution prepared by dissolving 37.5g citric acid, a triprotic acid with molar mass

Question

of 192.14g, in water to make 250mL solution: A. 0.780N; B. 0.147N; C. 0.288N; D. 0.0865N; E. 2.35N (how did you calculate this?)

Answer 1

M= mole/liter   N=  equivalents / liter    AS citric acid has three hydrogens that can participate in an acid /base reaction

then each mole of citric acid has 3 equivalents.

37.5 grams/192.14 grams/mole= 0.1952 moles

0.1952 moles /0.25 liters = 0.781 M since each mole has three equivalents the N is 3 X M = 2.34 N

Proof N = equivalents per liter .. an equivalent is that amount of material that provides or reacts with 1 mole of protons  sooo if N = 2.34  then that would be 2.34 Equivalents /liter

1 equivalent is 192.14/3 = 64grams  soo 2.34equivalents X 64 = 149.8 grams/liter       37.5 grams /250 mL = 150 grams /liter

Answer 2

ok so the number of moles is weight/RMM = 37.5/192.14

molarity is the number of moles in 1000 ml so as you have got but 250 ml molarity =4*37.5/192.14 =0.7806 m

Now you have a triprotic acid so its normality would be 3 times that value. I'm pretty sure it would be difficult to pull 3 protons off a weak acid like citric, but anyway the noramality is 0.2342.

Edit oops I meant 2.342 N