Sagot :
Solution:
Note: H₃PO₄ is the solute and water is the solvent.
Step 1: Calculate the mass of solvent.
Note: The density of water is 1 g/mL.
[tex]\text{mass}_{solvent} = \text{density}_{solvent} × \text{volume}_{solvent}[/tex]
[tex]\text{mass}_{solvent} = \text{1 g/m L} × \text{80 m L}[/tex]
[tex]\text{mass}_{solvent} = \text{80 g}[/tex]
Step 2: Calculate the mass of solution.
[tex]\text{mass}_{solution} = \text{mass}_{solute} + \text{mass}_{solvent}[/tex]
[tex]\text{mass}_{solution} = \text{30 g} + \text{80 g}[/tex]
[tex]\text{mass}_{solution} = \text{110 g}[/tex]
Step 3: Calculate the volume of solution.
[tex]\text{volume}_{solution} = \frac{\text{mass}_{solution}}{\text{density}_{solution}}[/tex]
[tex]\text{volume}_{solution} = \frac{\text{110 g}}{\text{1.88 g/m L}}[/tex]
[tex]\text{volume}_{solution} = \text{58.5 m L}[/tex]
[tex]\text{volume}_{solution} = \text{0.0585 L}[/tex]
Step 4: Calculate the molar mass of solute.
molar mass = (1.008 g/mol × 3) + (30.97 g/mol × 1) + (16.00 g/mol × 4)
molar mass = 97.99 g/mol
Step 5: Determine the number of replaceable ions of H₃PO₄ (solute).
Since the given solute is an acid, the number of replaceable ions is equal to the number of [tex]H^{+}[/tex] ions. H₃PO₄ has 3 replaceable [tex]H^{+}[/tex] ions.
Therefore,
no. of replaceable ions = 3 eq/mol
Step 6: Calculate the equivalent weight.
[tex]\text{equivalent weight} = \frac{\text{molar mass}}{\text{no. of replaceable ions}}[/tex]
[tex]\text{equivalent weight} = \frac{\text{97.99 g/mol}}{\text{3 eq/mol}}[/tex]
[tex]\text{equivalent weight = 32.663 g/eq}[/tex]
Step 7: Calculate the no. of equivalents of solute.
[tex]\text{equivalents of solute} = \frac{\text{mass}_{solute}}{\text{equivalent weight}}[/tex]
[tex]\text{equivalents of solute} = \frac{\text{30 g}}{\text{32.663 g/eq}}[/tex]
[tex]\text{equivalents of solute = 0.918 eq}[/tex]
Step 8: Calculate the normality of the solution.
[tex]\text{normality} = \frac{\text{equivalents of solute}}{\text{volume of solution}}[/tex]
[tex]\text{normality} = \frac{\text{0.918 eq}}{\text{0.0585 L}}[/tex]
[tex]\boxed{\text{normality = 15.7 N}}[/tex]
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