calculate the percent w/v, mole fraction of components, molarity and normality of the solution containing 13 g nahso4 dissolved in water upto total volume of 250 ml
To calculate the percent w/v, mole fraction, molarity, and normality of a solution, we need to follow a step-by-step process.
Step 1: Convert the given mass of NaHSO4 to grams.
Given: Mass of NaHSO4 = 13 g
Step 2: Convert the volume of the solution to liters.
Given: Volume of the solution = 250 ml = 250/1000 = 0.25 L
Now, let's calculate each parameter one by one:
1. Percent w/v:
Percent w/v = (mass of solute/volume of solution) x 100
Substituting the values:
Percent w/v = (13 g / 0.25 L) x 100
Percent w/v = 52%
2. Mole Fraction:
Mole fraction of a component X in a solution is given by the formula:
Mole fraction, X = (moles of X / total moles of all components)
To calculate the mole fraction of NaHSO4, we need to know the number of moles. To find the moles, we can use the molar mass of NaHSO4, which is as follows:
Molar mass of Na = 22.99 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16.00 g/mol
Adding these masses:
Molar mass of NaHSO4 = 22.99 + 1.01 + 32.07 + (4 × 16.00) = 120.06 g/mol
Now, let's find the moles of NaHSO4:
Moles of NaHSO4 = Mass of NaHSO4 / Molar mass
Moles of NaHSO4 = 13 g / 120.06 g/mol
Now, substituting the values into the mole fraction formula:
Mole fraction of NaHSO4 = (13 g / 120.06 g/mol) / [(13 g / 120.06 g/mol) + (0.25 g / 18.02 g/mol)]
Mole fraction of NaHSO4 = 0.163
3. Molarity:
Molarity is defined as the number of moles of solute per liter of solution.
Molarity (M) = Moles of solute / Volume of solution (in liters)
Now, substituting the values into the molarity formula:
Molarity (M) = (13 g / 120.06 g/mol) / 0.25 L
Molarity (M) = 0.434 M
4. Normality:
Normality is the number of gram equivalent weights of solute present in one liter of solution.
Normality (N) = (Equivalent weight of solute / Volume of solution (in liters))
The equivalent weight of NaHSO4 can be calculated as follows:
Equivalent weight of NaHSO4 = Molar weight of NaHSO4 / Number of equivalents
The number of equivalents of NaHSO4 is the same as its number of acidic hydrogen ions, which is 2.
Equivalent weight of NaHSO4 = Molar weight of NaHSO4 / 2
Now, substituting the values into the normality formula:
Normality (N) = (13 g / [(120.06 g/mol) / 2]) / 0.25 L
Normality (N) = 0.868 N
Therefore, the calculations give us the following results:
- Percent w/v = 52%
- Mole fraction of NaHSO4 = 0.163
- Molarity = 0.434 M
- Normality = 0.868 N