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.
