Question

How many grams of potassium nitrate are required to prepare 3.00 x 10^2 mL of 0.750 M solution

Answer 1

Mass = 22.725 g

Step-by-step explanation:

Given data:

Mass of potassium nitrate required = ?

Volume of solution = 3.00 ×10² mL

Molarity of solution = 0.750 M

Solution:

Formula:

Molarity = number of moles of solute / L of solution

Now we will convert the volume into L.

3.00 ×10² mL × 1L /1000 mL

0.003 ×10² L or 0.3 L

now we will put the values in formula.

0.750 M = n / 0.3 L

n = 0.750 M × 0.3 L

M = mol/L

n = 0.225 mol

Mass of potassium nitrate:

Mass = number of moles × molar mass

Mass = 0.225 mol × 101 g/mol

Mass = 22.725 g