A student weighs out 0.477g of Mg(NO3)2 and dissolves it in enough water to make 175ml of solution. She then pipettes exactly 10.00ml of the solution into a different vol. flask and adds enoguh water to make exactly 500.00ml of solution. What is the concentration of nitrate ion in the final solution?
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2018-03-20 11:55 am
回答 (1)
2018-03-20 3:50 pm
Step 1 : Dissolve 0.477 g Mg(NO₃)₂ to make 175 ml solution
Molar mass of Mg(NO₃)₂ = (24.3 + 14.0×2 + 16.0×6) g/mol = 148.3 g/mol
No. of moles of Mg(NO₃)₂ = (0.477 g) / (148.3 g/mol) = 0.003216 mol
Volume of the solution = 175 ml = 0.175 L
Molarity of Mg(NO₃)₂ = (0.003216 mol) / (0.175 L) = 0.01838 M
Step 2 : Dilute 10 ml of the above solution to 500.00 ml
C₁ = 0.01838n M, V₁ = 10.00 ml, C₂ = ? M, V₂ = 500.00 ml
C₁ V₁ = C₂ V₂
Final molarity of Mg(NO₃)₂, C₂ = C₁ × (V₁/V₂) = (0.01838 M) × (10.00/500.00) = 3.876 × 10⁻⁴ M
Each mole of Mg(NO₃)₂ contains 2 moles of NO₃²⁻ ion.
Concentration of NO₃²⁻ ion in the final solution = (3.876 × 10⁻⁴ M) × 2 = 7.35 × 10⁻⁴ M (to 3 sig. fig.)
Molar mass of Mg(NO₃)₂ = (24.3 + 14.0×2 + 16.0×6) g/mol = 148.3 g/mol
No. of moles of Mg(NO₃)₂ = (0.477 g) / (148.3 g/mol) = 0.003216 mol
Volume of the solution = 175 ml = 0.175 L
Molarity of Mg(NO₃)₂ = (0.003216 mol) / (0.175 L) = 0.01838 M
Step 2 : Dilute 10 ml of the above solution to 500.00 ml
C₁ = 0.01838n M, V₁ = 10.00 ml, C₂ = ? M, V₂ = 500.00 ml
C₁ V₁ = C₂ V₂
Final molarity of Mg(NO₃)₂, C₂ = C₁ × (V₁/V₂) = (0.01838 M) × (10.00/500.00) = 3.876 × 10⁻⁴ M
Each mole of Mg(NO₃)₂ contains 2 moles of NO₃²⁻ ion.
Concentration of NO₃²⁻ ion in the final solution = (3.876 × 10⁻⁴ M) × 2 = 7.35 × 10⁻⁴ M (to 3 sig. fig.)
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