If the unknown tri protic acid has an equivalent weight of 46.79 g/equiv, its average molecular weight is ?g/mol and ?g would be required to make 100.00 mL of a 1.50 N solution.
equivalent weight = molar mass/#H; therefore, I assume from the problem that all three H^+ were neutralized. Thus the molar mass s 3 x 46.79 = ??
N = equivalents/L
You know N = 1.50, and you know volume (0.100 L), calculate # equivalents and from that the mass of that number of equivalents.