Question

Calculate the molarity of each of the following solutions. Part A) 0.12 mol of LiNO3 in 5.5 L of solution Part B) 60.7 g C2H6O in 2.48 L of solution Part C) 14.2 mg KI in 100 mL of solution

Answer 1

For A: The molarity of solution is 0.218 M.

For B: The molarity of solution is 0.532 M.

For C: The molarity of solution is
`8.552* 10^(-4)M`

Step-by-step explanation:

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}`

Or,

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}`

• For A: 0.12 mol of
  `LiNO_3` in 5.5 L of solution

We are given:

Moles of
`LiNO_3` = 0.12 moles

Volume of the solution = 5.5 L

Putting values in above equation, we get:

`\text{Molarity of }LiNO_3=(0.12)/(5.5L)\\\\\text{Molarity of }LiNO_3}=0.0218M`

Hence, the molarity of solution is 0.0218 M.

• For B: 60.7 g
  `C_2H_6O` in 2.48 L of solution

We are given:

Given mass of
`C_2H_6O` = 60.7 g

Molar mass of
`C_2H_6O` = 46 g/mol

Volume of the solution = 2.48 L

Putting values in above equation, we get:

`\text{Molarity of }C_2H_6O=(60.7g)/(46g/mol* 5.5L)\\\\\text{Molarity of }C_2H_6O}=0.532M`

Hence, the molarity of solution is 0.532 M.

• For C: 14.2 mg KI in 100 mL of solution

We are given:

Given mass of KI = 14.2 mg =
`14.2* 10^(-3)g` (Conversion factor:
`1mg=10^(-3)g`

Molar mass of KI = 166 g/mol

Volume of the solution = 100 L

Putting values in above equation, we get:

`\text{Molarity of KI}=(14.2* 10^(-3)g* 1000)/(166g/mol* 100mL)\\\\\text{Molarity of KI}=8.552* 10^(-4)M`

Hence, the molarity of solution is
`8.552* 10^(-4)M`