what final volume in liters of a 2.5 m solution of sodium nitrate can be made using 25.0g of solution
To find the final volume in liters of a 2.5 m solution of sodium nitrate using 25.0g of sodium nitrate, you need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
1. First, you need to calculate the number of moles of sodium nitrate using its molar mass.
The molar mass of sodium nitrate (NaNO3) can be calculated by summing the atomic masses of all the elements in it.
Molar mass of NaNO3 = (22.99 g/mol) + (14.01 g/mol) + (3 * 16.00 g/mol) = 85.00 g/mol
2. Calculate the number of moles of sodium nitrate:
moles = mass / molar mass
moles = 25.0 g / 85.00 g/mol
3. Now, use the molarity formula to find the volume of the solution:
2.5 M = moles / volume
Rearranging the formula, we get:
volume = moles / 2.5 M
Substituting the values:
volume = (25.0 g / 85.00 g/mol) / 2.5 M
4. Calculate the final volume in liters:
Convert grams to moles by dividing by the molar mass:
moles = 25.0 g / 85.00 g/mol
Substitute the moles value into the volume formula:
volume = (25.0 g / 85.00 g/mol) / 2.5 M
5. Calculate the final volume:
volume = 0.294 L or 294 mL (rounded to three decimal places)
Therefore, the final volume of the 2.5 M sodium nitrate solution that can be made using 25.0 g is approximately 0.294 L.