If the unknown tri protic acid has an equivalent weight of 139.42 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.
See my response to your first question.
To find the average molecular weight of the unknown triprotic acid, we need to determine the number of equivalents in one mole of the acid. Since the equivalent weight is given as 139.42 g/equiv, we can calculate the average molecular weight as follows:
Average molecular weight (g/mol) = Equivalent weight (g/equiv) * Number of equivalents (equiv/mol)
To find the number of equivalents in one mole of the acid, we need to determine the acid's equivalent weight. Since the acid is triprotic, it can donate three protons or equivalents per molecule.
Number of equivalents (equiv/mol) = 3
Therefore, the average molecular weight of the acid is:
Average molecular weight (g/mol) = 139.42 g/equiv * 3 equiv/mol = 418.26 g/mol
To calculate the amount of acid required to make a 1.50 N solution in a 100.00 mL volume, we need to first convert the solution concentration from N (normality) to molarity (mol/L).
N = (mol of solute) / (L of solution)
To convert N to molarity, we need to know the number of equivalents of solute in the reaction. In this case, since the acid is triprotic, we have three equivalents.
1.50 N = (mol of acid) / (0.100 L)
To find the moles of acid required, we rearrange the equation:
mol of acid = 1.50 N * 0.100 L = 0.150 mol
Now, we can calculate the mass of acid required using the average molecular weight:
Mass of acid (g) = moles of acid * average molecular weight (g/mol)
Mass of acid (g) = 0.150 mol * 418.26 g/mol = 62.739 g
Therefore, 62.739 g of the unknown triprotic acid would be required to make 100.00 mL of a 1.50 N solution.