Questions on Solutions and Dilutions help! I have the answer just not the process!!?

[匿名]

2020-11-24 2:49 pm

1) If [K₃PO₄ ] = 0.250 M, how many grams of K₃PO₄ are in 750.0 mL of solution? Answer: 39.8g K₃PO₄

2) What volume in mL of 0.600 M aluminum sulfate solution can be prepared using 35.00 grams of aluminum sulfate ? Answer: 170. mL Al₂(SO₄)₃

3) What is the resulting concentration of BaBr₂₍aq₎ when 400. mL of water are added to 375. mL of 0.350 M BaBr₂(aq)? Answer: 0.169 M

4) How much water needs to be added to 15.0 mL of a concentrated solution of HCl(aq) to lower its concentration to 1.00 M? [concentrated HCl(aq)] =11.6 M. Answer: Water added = 159 mL

回答 (1)

Uncle Michael

2020-11-24 3:42 pm

✔ 最佳答案

1)

Molar mass of K₃PO₄ = (39.1×3 + 31.0 + 16.0×4) g/mol = 212.3 g/mol

(0.250 mol K₃PO₄ / 1000 mL solution) × (750.0 mL solution) × (212.3 g K₃PO₄ / 1 mol K₃PO₄)
= 39.8 g K₃PO₄

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2)
Molar mass of Al₂(SO₄)₃ = (27×2 + 32.1×3 + 16.0×12) g/mol = 342.3 g/mol

[35.00 g Al₂(SO₄)₃] × [1 mol Al₂(SO₄)₃ / 342.3 g Al₂(SO₄)₃] × [1000 mL Al₂(SO₄)₃ / 0.600 mol]
= 170. mL Al₂(SO₄)₃

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3)
For dilution: C₁V₁ = C₂V₂
Then, C₂ = C₁ × [V₁/V₂]

Final concentration, C₂ = (0.350 M) × [375/(375 + 400)] = 0.169 M

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4)
For dilution: C₁V₁ = C₂V₂
Then, V₂ = V₁ × (C₁/C₂)

Final volume, V₂ = (15.0 mL) × (11.6/1.00) = 174 mL
Volume of water added = (174 - 15.0) mL = 159 mL

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