Which of the following solutions has a molarity different from the rest?
a. 6 N H3PO4
b. 4 N Ca(OH)2
c. 8 N HC2H3O2
d. 2 N NH40H
N = #H or OH*M or
M = N/#H or OH
For H3PO4. 6N/3 = 2M
For Ca(OH)2. 4N/2 = 2
etc.
M =
Well, it seems like these solutions are having a little "acidic" competition. But don't worry, I'm here to help you out with a dash of humor!
Out of all the options, the oddball is d. 2 N NH40H, who found themselves in a "basic" mix-up with the other acidic solutions. Poor little NH40H just couldn't keep up with the "N" crowd.
To determine which solution has a molarity different from the rest, we need to compare the molarities of the given solutions.
The molarity of a solution is defined as the number of moles of solute divided by the volume of the solution in liters.
Let's calculate the molarities of the given solutions:
a. 6 N H3PO4:
The subscript 'N' indicates normality, which is a measure of solution concentration. Normality is the number of gram equivalent weights of solute per liter of solution.
To calculate the molarity from normality, we need to know the equivalent weight of the solute. The equivalent weight is the molar mass of the compound divided by its valence.
The molar mass of H3PO4 can be calculated as:
(1 * 1.01 g/mol) + (3 * 1.01 g/mol) + (1 * 16.00 g/mol) + (4 * 16.00 g/mol) = 98.00 g/mol
The valence of H3PO4 is 3 because it has three hydrogen ions (H+) in the formula.
Thus, the equivalent weight of H3PO4 is 98.00 g/mol / 3 = 32.67 g/mol.
The molarity can be calculated from normality by multiplying by the equivalent weight:
6 N * 32.67 g/mol = 196.02 g/L
So, the molarity of 6 N H3PO4 is 196.02 M.
b. 4 N Ca(OH)2:
Using the same logic as above, we can calculate the equivalent weight of Ca(OH)2.
The molar mass of Ca(OH)2 can be calculated as:
(1 * 40.08 g/mol) + (2 * (1 * 1.01 g/mol + 1 * 16.00 g/mol)) = 74.10 g/mol
The valence of Ca(OH)2 is 2 because it has two hydroxide ions (OH-) in the formula.
Thus, the equivalent weight of Ca(OH)2 is 74.10 g/mol / 2 = 37.05 g/mol.
The molarity can be calculated from normality by multiplying by the equivalent weight:
4 N * 37.05 g/mol = 148.20 g/L
So, the molarity of 4 N Ca(OH)2 is 148.20 M.
c. 8 N HC2H3O2:
Using the same logic as above, we can calculate the equivalent weight of HC2H3O2.
The molar mass of HC2H3O2 can be calculated as:
(1 * 1.01 g/mol) + (2 * 12.01 g/mol) + (2 * 1.01 g/mol) + (2 * 16.00 g/mol) = 60.05 g/mol
The valence of HC2H3O2 is 1 because it has one hydrogen ion (H+) in the formula.
Thus, the equivalent weight of HC2H3O2 is 60.05 g/mol / 1 = 60.05 g/mol.
The molarity can be calculated from normality by multiplying by the equivalent weight:
8 N * 60.05 g/mol = 480.40 g/L
So, the molarity of 8 N HC2H3O2 is 480.40 M.
d. 2 N NH40H:
Using the same logic as above, we can calculate the equivalent weight of NH40H.
The molar mass of NH40H can be calculated as:
(1 * 14.01 g/mol) + (4 * 1.01 g/mol) + (1 * 16.00 g/mol) = 35.05 g/mol
The valence of NH40H is 1 because it has one ammonium ion (NH4+) in the formula.
Thus, the equivalent weight of NH40H is 35.05 g/mol / 1 = 35.05 g/mol.
The molarity can be calculated from normality by multiplying by the equivalent weight:
2 N * 35.05 g/mol = 70.10 g/L
So, the molarity of 2 N NH40H is 70.10 M.
Comparing the molarities of all the solutions:
- 6 N H3PO4: 196.02 M
- 4 N Ca(OH)2: 148.20 M
- 8 N HC2H3O2: 480.40 M
- 2 N NH40H: 70.10 M
From the calculations, we can see that the molarity of 8 N HC2H3O2 is significantly higher than the molarities of the other solutions. Therefore, option c, 8 N HC2H3O2, has a molarity different from the rest.
To determine which solution has a different molarity from the rest, we need to compare the values of the molarity for each solution. Molarity represents the number of moles of solute dissolved in one liter of solution.
To calculate the molarity of a solution, we need to know the number of moles of solute and the volume of the solution in liters. The formula for molarity is given as:
Molarity (M) = Moles of Solute (mol) / Volume of Solution (L)
Let's calculate the molarities for each solution:
a. 6 N H3PO4:
"N" refers to normality, which is the concentration of a solution in terms of equivalents per liter. Since H3PO4 is a triprotic acid, 6 N means that it contains three times more equivalents of acid per liter. To convert it to molarity, we need to divide by the number of equivalents per mole of the acid. H3PO4 has three equivalents per mole.
Molarity = Normality / Number of Equivalents per Mole
Molarity = (6 N) / 3
Molarity = 2 M
b. 4 N Ca(OH)2:
Similar to the previous example, "N" represents normality. Ca(OH)2 is a strong base that contains two equivalents of hydroxide ions per mole.
Molarity = Normality / Number of Equivalents per Mole
Molarity = (4 N) / 2
Molarity = 2 M
c. 8 N HC2H3O2:
Again, "N" stands for normality. HC2H3O2 is a weak acid that contains one equivalent of acid per mole.
Molarity = Normality / Number of Equivalents per Mole
Molarity = (8 N) / 1
Molarity = 8 M
d. 2 N NH4OH:
Once more, "N" denotes normality. NH4OH is a weak base that contains one equivalent of hydroxide ions per mole.
Molarity = Normality / Number of Equivalents per Mole
Molarity = (2 N) / 1
Molarity = 2 M
After calculating the molarities, we find that each solution has a molarity of 2 M, except for solution c. The solution with a different molarity from the rest is option c.