Question

State the mass of the solute required to prepare 4.00 L of 0.55 M Hg(NO3)2 solution.

Answer 1

714.34 grams

Step-by-step explanation

Given:

Volume of the solution, V = 4.00 L

Molarity of the solution =0.55 M

What to find:

The mass of the solute required to make the solution.

Step-by-step solution:

Step 1: Calculate the mole of the solute.

The moles of the solute can be determined using the molarity formula:

`Molarity=\frac{Moles}{Volume\text{ }in\text{ }liters}`

Plugging the values of the parameters into the formula:

`\begin{gathered} 0.55\text{ }M=\frac{Moles}{4.00\text{ }L} \\ \\ \Rightarrow Moles=0.55\text{ }M*4.00\text{ }L \\ \\ Moles=2.2\text{ }mol \end{gathered}`

Step 2: Convert the moles of solute to the mass of solute:

Using the mole formula, the mass of the solute can be calculated as follows:

`Moles=\frac{Mass}{Molar\text{ }mass}`

Using the atomic mass of each element in the periodic table, the molar mass of Hg(NO3)2 = 324.7 g/mol.

Putting mole = 2.2 mol and molar mass = 324.7 g/mol, the mass of the solute is:

`\begin{gathered} 2.2mol=\frac{Mass}{324.7g\text{/}mol} \\ \\ \Rightarrow Mass=2.2mol*324.7g\text{/}mol \\ \\ Mass=714.34\text{ }g \end{gathered}`

Therefore, the solute required to prepare 4.00 L of 0.55 M Hg(NO3)2 solution = 714.34 grams.