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

Answer:

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 $8.552\times 10^{-4}M$

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,

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

• 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:

$\text{Molarity of }LiNO_3=\frac{0.12}{5.5L}\\\\\text{Molarity of }LiNO_3}=0.0218M$

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:

$\text{Molarity of }C_2H_6O=\frac{60.7g}{46g/mol\times 5.5L}\\\\\text{Molarity of }C_2H_6O}=0.532M$

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 = $14.2\times 10^{-3}g$     (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:

$\text{Molarity of KI}=\frac{14.2\times 10^{-3}g\times 1000}{166g/mol\times 100mL}\\\\\text{Molarity of KI}=8.552\times 10^{-4}M$

Hence, the molarity of solution is $8.552\times 10^{-4}M$