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boyakko [2]
3 years ago
15

Calculate the molar mass of Ca3(PO4)2 934.32 g Ca3(PO4)2 =______ moles Ca3(PO4)2

Chemistry
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

  • Molar mass of Ca₃(PO₄)₂ = 310.18 g/mol
  • 934.32 g Ca₃(PO₄)₂ = 3.01 moles Ca₃(PO₄)₂

Explanation:

The<em> molar mass </em>(MM)<em> of Ca₃(PO₄)₂</em> can be calculated as follows:

  • MM of Ca₃(PO₄)₂ = (MM of Ca)*3 + [(MM of P) +(MM of O)*4]*2
  • MM of Ca₃(PO₄)₂ = 310.18 g/mol

Now we can <u>convert 934.32 g of Ca₃(PO₄)₂ into moles</u>:

  • 934.32 g ÷ 310.18 g/mol = 3.01 moles

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Read 2 more answers
What is the mass percent of sucrose (C12H22O11, Mm = 342 g/mol) in a 0.329-m sucrose solution?
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Answer:

\% m/m=10.1\%

Explanation:

Hello,

In this case given the molal solution of sucrose, we can assume there are 0.329 moles of sucrose in 1 kg of solvent, thus, computing both the mass of sucrose and solvent in grams, we obtain:

m_{sucrose}=0.329mol*\frac{342g}{1mol}=112.5g

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\% m/m=\frac{112.5g}{112.5g+1000g}*100\%\\ \\\% m/m=10.1\%

Regards.

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