Question

Describe how you would prepare 250 mL of a 0.707 M NaNO3 solution.

Answer 1

Answer:

To prepare 250 mL of a 0.707 M NaNO3 solution, take 15.045 g of NaNO₃ in 250 mL flask and fill with solvent up to mark.

Explanation:

Data given:

volume of solution = 250 mL

Convert mL to L

1 ml = 1000 L

250 ml = 250/1000 = 0.25

Molarity of solution = 0.707 M

How to prepare = ?

Solution:

Molarity:

This the term used for the concentration of the solution. It is the amount of solute in moles dissolve in 1 Liter of solution.

So we have to calculate amount of NaNO₃

So,

we have to know the number of moles of solution that will be required

Formula used for Molarity

         Molarity = number of moles of solute / L of solution  

Rearrange above equation

        number of moles of solute = Molarity x L of solution . . . . . . . (1)

Put values in equation 1

        number of moles of solute = 0.707 mol/L x 0.25 L

        number of moles of solute = 0.177 mol

So we need 0.177 mol of NaNO₃ to prepare 250 mL of a 0.707 M solution.

Now convert moles to mass

            Mass = no. of moles x Molar mass . . . . . . . . . . (2)

molar mass of NaNO₃ = 23 + 14 + 3(16)

molar mass of NaNO₃ = 85 g/mol

Put value in equation 2

          Mass of NaNO₃= 0.177 mol x 85 g/mol

          Mass of NaNO₃=  15.045 g

So now,

To prepare 250 mL of a 0.707 M NaNO3 solution, take 15.045 g of NaNO₃ in 250 mL flask and fill with solvent up to mark.