Answer:
Step-by-step explanation:
For NaOH:
Molarity = moles of solute / liters of solution
Molality = moles of solute / mass of solvent (in kg)
Normality = equivalents of solute / liters of solution
First, we need to find the moles of NaOH:
moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 20 g / 40 g/mol
moles of NaOH = 0.5 mol
Next, we need to find the liters of solution:
liters of solution = 600 mL / 1000 mL/L
liters of solution = 0.6 L
Now we can calculate the molarity:
Molarity = 0.5 mol / 0.6 L
Molarity = 0.833 M
To find the molality, we need to find the mass of the solvent (water) in kg:
mass of solvent = volume of solution x density of solvent
mass of solvent = 0.6 L x 1.10 g/mL
mass of solvent = 0.66 kg
Now we can calculate the molality:
Molality = 0.5 mol / 0.66 kg
Molality = 0.758 mol/kg
Finally, to find the normality, we need to know that NaOH is a monoprotic base (i.e. donates one H+ ion per molecule). Therefore, the equivalents of NaOH is equal to the moles of NaOH:
Normality = 0.5 eq / 0.6 L
Normality = 0.833 eq/L
For Ba(OH)2:
Molarity = moles of solute / liters of solution
Molality = moles of solute / mass of solvent (in kg)
Normality = equivalents of solute / liters of solution
First, we need to find the moles of Ba(OH)2:
moles of Ba(OH)2 = mass of Ba(OH)2 / molar mass of Ba(OH)2
moles of Ba(OH)2 = 50 g / (137.33 g/mol x 2)
moles of Ba(OH)2 = 0.182 mol
Next, we need to find the liters of solution:
liters of solution = 800 mL / 1000 mL/L
liters of solution = 0.8 L
Now we can calculate the molarity:
Molarity = 0.182 mol / 0.8 L
Molarity = 0.227 M
To find the molality, we need to find the mass of the solvent (water) in kg:
mass of solvent = volume of solution x density of solvent
mass of solvent = 0.8 L x 1.15 g/mL
mass of solvent = 0.92 kg
Now we can calculate the molality:
Molality = 0.182 mol / 0.92 kg
Molality = 0.198 mol/kg
Finally, to find the normality, we need to know that Ba(OH)2 is a diprotic base (i.e. donates two H+ ions per molecule). Therefore, the equivalents of Ba(OH)2 is twice the moles of Ba(OH)2:
Normality = 0.364 eq / 0.8 L
Normality = 0.455 eq/L
Molarity of H3PO4:
First, calculate the number of moles of H3PO4:
m(H3PO4) = 40 g
Molecular weight of H3PO4 = 3(1.01) + 1.01 + 4(16) = 98 g/mol
n(H3PO4) = m(H3PO4) / M(H3PO4) = 40 g / 98 g/mol = 0.4082 mol
Now, calculate the volume of the solution in liters:
V = 900 mL = 0.9 L
Molarity (M) = n / V = 0.4082 mol / 0.9 L = 0.4536 M
Molality of H3PO4:
Mass of solvent (water) = (900 mL) x (1.30 g/mL) - 40 g = 1040 g
Molality (m) = n(H3PO4) / m(solvent) = 0.4082 mol / 1.040 kg = 0.3929 m
Normality of H3PO4:
H3PO4 is a triprotic acid, meaning that it can donate up to three protons (H+ ions) per molecule. The normality of H3PO4 depends on which proton(s) we are interested in. Here, we will calculate the normality with respect to the first proton.
One mole of H3PO4 can donate three moles of H+ ions, so the equivalent weight (the weight that can donate one mole of H+ ions) of H3PO4 is its molecular weight divided by three:
Equiv. Wt. = 98 g/mol / 3 = 32.67 g/mol
Normality (N) = Molarity x Number of H+ ions donated per molecule / Equivalent weight
For the first proton, the number of H+ ions donated per molecule is 1. Therefore:
N = 0.4536 M x 1 / 32.67 g/mol = 0.01388 N
First, we need to calculate the number of moles of Na2CO3:
Molar mass of Na2CO3 = 2(23) + 12 + 3(16) = 106 g/mol
Number of moles of Na2CO3 = mass/molar mass = 25 g / 106 g/mol = 0.235 moles
Next, we can calculate the molarity, molality and normality using the following formulas:
Molarity = moles of solute / liters of solution
Molarity = 0.235 moles / 1 L = 0.235 M
Molality = moles of solute / mass of solvent (in kg)
Density of solution = 1.05 g/mL = 1050 kg/m³
Volume of solution = 1000 mL = 1 L
Mass of solvent = volume of solution x density of solution = 1 L x 1050 kg/m³ = 1050 g = 1.05 kg
Molality = 0.235 moles / 1.05 kg = 0.224 mol/kg
Normality = (moles of solute x equivalent weight) / liters of solution
The equivalent weight of Na2CO3 is equal to its molar mass divided by the number of hydrogen ions it can donate in a reaction, which is 2.
Equivalent weight of Na2CO3 = molar mass / 2 = 106 g/mol / 2 = 53 g/eq
Normality = (0.235 moles x 53 g/eq) / 1 L = 12.455 N