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Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M
Answer:
Mass = 153.48 g
Explanation:
Given data:
Volume of solution = 2.50 L
Molarity = 0.48 M
Mass required = ?
Solution:
Molarity = number of moles / volume in litter
Number of moles = Molarity × volume in litter
Number of moles = 0.48 M × 2.50 L
Number of moles = 1.2 mol
Mass of HI:
Number of moles = mass/molar mass
Mass = Number of moles × molar mass
Mass = 1.2 mol × 127.9 g/mol
Mass = 153.48 g
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