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Sagot :
Answer
979.94 grams
Explanation
Given:
Volume of solution = 2 L
Molarity = 5 M
What to find:
The mass in grams of H3PO4 present in the solution.
Step-by-step solution:
The first step is to calculate the moles of H3PO4 present using the molarity formula below:
[tex]\begin{gathered} Molarity=\frac{Moles}{Volume\text{ }in\text{ }L} \\ \\ \Rightarrow Moles=Molarity\times Volume\text{ }in\text{ }L \end{gathered}[/tex]Put the values of the given parameters into the formula to get the moles:
[tex]Moles=5M\times2L=10\text{ }mol[/tex]The moles of H3PO4 present in the solution = 10 mol.
Therefore, the mass present can be determined using the mole formula:
[tex]\begin{gathered} Moles=\frac{Mass\text{ }in\text{ }grams}{Molar\text{ }mass} \\ \\ \Rightarrow Mass\text{ }in\text{ }grams=Moles\times Molar\text{ }mass \end{gathered}[/tex]From the periodic table, the molar mass of H3PO4 can be known as 97.994 g/mol.
So putting moles = 10 mol and molar mass = 97.994 g/mol, then the mass is:
[tex]Mass=10\text{ }mol\times97.994\text{ }g\text{/}mol=979.94\text{ }grams[/tex]Therefore, the mass in grams of H3PO4 present in the solution = 979.94 grams.
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