molarity of 0.6N aluminium sulphate is?
Normality=M×total +ve or -ve charge
To calculate the molarity of a solution, you need to know the concentration in moles per liter (mol/L) and the formula weight or molar mass of the solute.
In the case of aluminum sulfate (Al2(SO4)3), the molar mass can be calculated by adding up the atomic masses of all the atoms in the formula. The atomic mass of aluminum (Al) is 26.98 g/mol, and the atomic mass of sulfur (S) is 32.07 g/mol. The atomic mass of oxygen (O) is 16.00 g/mol, but there are four oxygen atoms in each molecule of aluminum sulfate. Therefore, the molar mass of aluminum sulfate can be calculated as follows:
2(26.98 g/mol) + 3(32.07 g/mol) + 4(16.00 g/mol) = 54.98 g/mol + 96.21 g/mol + 64.00 g/mol = 215.19 g/mol
Now, let's calculate the molarity using the given concentration of 0.6N (normality). Normality is defined as the number of equivalents (moles of solute) in one liter of solution.
To convert normality to molarity, we need to know the equivalent weight of the solute. For aluminum sulfate, the equivalent weight is half of the molar mass because each mole of aluminum sulfate produces two equivalents of aluminum ions. Therefore, the equivalent weight of aluminum sulfate is 215.19 g/mol / 2 equivalents = 107.60 g/mol or 0.1076 kg/mol.
Now let's calculate the molarity:
Given: Normality (N) = 0.6 N
Equivalent weight (E) = 0.1076 kg/mol
Molarity (M) = Normality (N) / Equivalent weight (E)
Molarity (M) = 0.6 N / 0.1076 kg/mol
To convert kg/mol to g/L, we need to multiply by 1000 (1 L = 1000 mL, 1 kg = 1000 g):
Molarity (M) = (0.6 N / 0.1076 kg/mol) * 1000 g/L
Molarity (M) = 5567.38 g/L
Therefore, the molarity of 0.6N aluminum sulfate is approximately 5567.38 M
koja
Technically one must know the reaction involved; however, in problems like this the usual way of answering is M = N x positive or negative charge. For Al2(SO4)3 that is 2*3 = 6; therefore,
M = N x valence = 0.6 x 6 = ?