Sagot :
Given:
mass of solute = 55 g
volume of solution = 1.4 L
solute: H₂SO₄
Required:
normality
Solution:
Step 1: Calculate the molar mass of solute.
molar mass of solute = (1.008 g/mol × 2) + (32.07 g/mol × 1) + (16.00 g/mol × 4)
molar mass of solute = 98.086 g/mol
Step 2: Determine the number of replaceable ions of the solute.
Since the given solute is H₂SO₄, which is an acid, the number of active species is equal to the number of H⁺ ions present. H₂SO₄ has two active H⁺ ions in 1 mole. Therefore,
number of active species = 2 eq/mol
Step 3: Calculate the equivalent weight of solute.
[tex]\text{equivalent weight of solute} = \frac{\text{molar mass of solute}}{\text{number of active species}}[/tex]
[tex]\text{equivalent weight of solute} = \frac{\text{98.086 g/mol}}{\text{2 eq/mol}}[/tex]
equivalent weight of solute = 49.043 g/eq
Step 4: Calculate the number of equivalents of solute.
[tex]\text{equivalents of solute} = \frac{\text{mass of solute}}{\text{equivalent weight of solute}}[/tex]
[tex]\text{equivalents of solute} = \frac{\text{55 g}}{\text{49.043 g/eq}}[/tex]
equivalents of solute = 1.121465 eq
Step 5: Calculate the normality of the solution.
[tex]\text{normality} = \frac{\text{equivalents of solute}}{\text{volume of solution}}[/tex]
[tex]\text{normality} = \frac{\text{1.121465 eq}}{\text{1.4 L}}[/tex]
[tex]\boxed{\text{normality = 0.80 N}}[/tex]
[tex]\\[/tex]
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