Question

5. You have 2 Liters of a 5 M Solution of H3PO4. How many grams are present?

Answer 1

979.94 grams

Step-by-step 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:

`\begin{gathered} Molarity=\frac{Moles}{Volume\text{ }in\text{ }L} \\ \\ \Rightarrow Moles=Molarity* Volume\text{ }in\text{ }L \end{gathered}`

Put the values of the given parameters into the formula to get the moles:

`Moles=5M*2L=10\text{ }mol`

The moles of H3PO4 present in the solution = 10 mol.

Therefore, the mass present can be determined using the mole formula:

`\begin{gathered} Moles=\frac{Mass\text{ }in\text{ }grams}{Molar\text{ }mass} \\ \\ \Rightarrow Mass\text{ }in\text{ }grams=Moles* Molar\text{ }mass \end{gathered}`

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:

`Mass=10\text{ }mol*97.994\text{ }g\text{/}mol=979.94\text{ }grams`

Therefore, the mass in grams of H3PO4 present in the solution = 979.94 grams.