BySky168
Hello Community,
I don't understand how to calculate this problem:
How many grams of 96% (m/m) sulfuric acid (Mw(H2SO4)= 98.1g/mol) must be dissolved in one liter to obtain a 1 normal sulfuric acid solution? Please round the result to three significant places.
How do you proceed here? What is a normal?
I would be happy to receive a detailed calculation.
Thanks in advance!😄
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Normality (unit characters) N) is an obsolete concentration measure of chemistry. It is currently no longer part of the DIN standards and was obtained by the equivalent concentration (unit character c)eq) replaced. Normality is the number of gram equivalents or val dissolved in one liter of a solution. A gram equivalent is that mass of substance which can release or absorb a mole of, for example, protons or electrons.
In the case of H2 SO4, which can release two protons, the normality N is twice the molarity M. ceq(H)2SO4= 2 * c(H)2SO4)
So we simply calculate and halve the molar mass concentration of sulfuric acid. Of course, exactly 1 mol in one liter of solution is required for a c=1 mol/L. These are 98.1 g. However, the proportion by mass of sulfuric acid is only 0.96 (96%). Therefore, the necessary mass is:
m = 98.1 g/0.96 = 102.2 g
102.2 g of the acid to one liter has the concentration:
c = 1 M (mol/L)
Thus, 51.1 g of acid per liter have the eqivalence concentration ceq(H)2SO4) = 1 N (Val/L)
Thank you.
To obtain a 1 normal sulfuric acid solution, a sulfuric acid concentration of 1 mol/L is required. The molar mass of sulfuric acid (H2 SO4) is 98.1 g/mol.
To calculate the amount of sulfuric acid required to produce 1 L of a 1 M solution, we first have to calculate the number of mols of sulfuric acid to be contained in 1 L:
1 mol/L = x mol/1 L
x = 1 mol
Therefore, we must have 1 mol of sulfuric acid in 1 L solution.
To calculate the amount of sulfuric acid in grams we need, we can use the formula for the amount of substance (n = m/M):
n=1 mol
M=98.1 g/mol
n = m/M/> m = n * M
m=1 mol*98.1 g/mol=98.1 g
Therefore, we need to dissolve 98.1 g of a 96% sulfuric acid (m/m) in 1 l of water to obtain a 1 normal sulfuric acid solution.
Isn’t that important?