Question

What mass of NaNO₃ is required to prepare 31.8 mL of a solution of NaNO₃ with a molarity of 0.138 M?

Answer 1

To prepare a solution of NaNO₃ with a molarity of 0.138 M, you will need 0.374 g of NaNO₃.

Step-by-step explanation:

To find the mass of NaNO₃ required, we need to use the formula for molarity:

Moles of solute = Molarity × Volume of solution

First, convert the given volume of solution to liters:

31.8 mL ÷ 1000 = 0.0318 L

Next, plug in the values into the formula:

Moles of NaNO₃ = 0.138 M × 0.0318 L = 0.0044 mol

Finally, use the molar mass of NaNO₃ (85.0 g/mol) to calculate the mass:

Mass of NaNO₃ = 0.0044 mol × 85.0 g/mol = 0.374 g