Question

How many grams of potassium benzoate trihydrate, KC H302 - 3H20. are needed to prepare 1.5 L of a 0.120M solution of potassium benzoate? Mass = g

Answer 1

Answer: The mass of potassium benzoate is $8.40\times 10^{-4}g$

Explanation:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Molarity of solution = 0.120 M

Molar mass of potassium benzoate trihydrate = 214.26 g/mol

Volume of solution = 1.5 L

Putting values in above equation, we get:

$0.120M=\frac{\text{Mass of potassium benzoate trihydrate}}{214.26\times 1.5}\\\\\text{Mass of potassium benzoate trihydrate}=\frac{0.120mol/L\times 1.5L}{214.26g/mol}=8.40\times 10^{-4}g$

Hence, the mass of potassium benzoate is $8.40\times 10^{-4}g$