Calculate the molarity and normality of a solution containing 9.8 g of H2SO4 on 250 cm cube of the solutiom
0.4
0.8 N
To calculate the molarity and normality of a solution, we need to know the molecular weight of the solute and the volume of the solution. In this case, we have the weight of the solute (9.8 g) and the volume (250 cm³) of the solution.
First, we need to calculate the number of moles of H2SO4 using its molecular weight. The molecular weight of H2SO4 is calculated as follows:
H (hydrogen) = 1 g/mol
S (sulfur) = 32 g/mol
O (oxygen) = 16 g/mol
Molecular weight of H2SO4 = (2 × 1) + 32 + (4 × 16) = 98 g/mol
The number of moles (n) can be calculated using the formula:
n = mass / molecular weight
n = 9.8 g / 98 g/mol = 0.1 mol
Next, we can calculate the molarity of the solution by dividing the number of moles by the volume in liters:
Molarity (M) = moles / volume (in liters)
Volume = 250 cm³ = 250/1000 = 0.25 L
Molarity (M) = 0.1 mol / 0.25 L = 0.4 mol/L
Now, to calculate the normality of the solution, we need to determine the number of equivalents of H2SO4. Since H2SO4 is a diprotic acid (can donate two protons), the number of equivalents is twice the number of moles.
Number of equivalents (N) = 2 × moles
Number of equivalents (N) = 2 × 0.1 mol = 0.2 N
Therefore, the molarity of the solution is 0.4 mol/L and the normality of the solution is 0.2 N.
0.05
I assume you meant to type, "9.8 g of H2SIO4 IN 250 cc of the soln."
mols H2SO4 = grams/molar mass
Then M = mols/L soln.
Technically you can't calculate the normality without knowing the equation in which it is to be used; however, usually when a question like this is asked it is ok to answer that N = M x number of replaceable H atoms. In this case there are two H atoms that can be replaced.