Respuesta :
Answer:
a) 0.125 M.
b) 0.04888 M.
c) 0.2056 M.
d) 5.625 x 10⁻³ M.
Explanation:
- To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.
(MV)before dilution = (MV)after dilution
(a) 1.00 L of a 0.250 M solution of Fe(NO₃)₃ is diluted to a final volume of 2.00 L.
∵ (MV)before dilution of Fe(NO₃)₃ = (MV)after dilution of Fe(NO₃)₃
M before dilution = 0.250 M, V before dilution = 1.00 L.
M after dilution = ??? M, V after dilution = 2.00 L.
∵ (MV)before dilution of Fe(NO₃)₃ = (MV)after dilution of Fe(NO₃)₃
∴ (0.250 M)(1.00 L) = (M after dilution of Fe(NO₃)₃)(2.00 L)
∴ M after dilution of Fe(NO₃)₃ = (0.250 M)(1.00 L)/(2.00 L) = 0.125 M.
(b) 0.5000 L of a 0.1222 M solution of C₃H₇OH is diluted to a final volume of 1.250 L.
∵ (MV)before dilution of C₃H₇OH = (MV)after dilution of C₃H₇OH
M before dilution = 0.1222 M, V before dilution = 0.500 L.
M after dilution = ??? M, V after dilution = 1.250 L.
∵ (MV)before dilution of C₃H₇OH = (MV)after dilution of C₃H₇OH
∴ (0.1222 M)(0.500 L) = (M after dilution of C₃H₇OH)(1.250 L)
∴ M after dilution of C₃H₇OH = (0.1222 M)(0.500 L)/(1.250 L) = 0.04888 M.
(c) 2.35 L of a 0.350 M solution of H₃PO₄ is diluted to a final volume of 4.00 L.
∵ (MV)before dilution of H₃PO₄ = (MV)after dilution of H₃PO₄
M before dilution = 0.350 M, V before dilution = 2.35 L.
M after dilution = ??? M, V after dilution = 4.00 L.
∵ (MV)before dilution of H₃PO₄ = (MV)after dilution of H₃PO₄
∴ (0.350 M)(2.35 L) = (M after dilution of H₃PO₄)(4.00 L)
∴ M after dilution of H₃PO₄ = (0.350 M)(2.35 L)/(4.00 L) = 0.2056 M.
(d) 22.50 mL of a 0.025 M solution of C₁₂H₂₂O₁₁ is diluted to 100.0 mL.
∵ (MV)before dilution of C₁₂H₂₂O₁₁ = (MV)after dilution of C₁₂H₂₂O₁₁
M before dilution = 0.025 M, V before dilution = 22.50 mL.
M after dilution = ??? M, V after dilution = 100.0 mL.
∵ (MV)before dilution of C₁₂H₂₂O₁₁ = (MV)after dilution of C₁₂H₂₂O₁₁
∴ (0.025 M)(22.50 mL) = (M after dilution of C₁₂H₂₂O₁₁)(100.0 mL)
∴ M after dilution of C₁₂H₂₂O₁₁ = (0.025 M)(22.50 mL)/(100.0 L) = 5.625 x 10⁻³ M.