if you want a solution that is .1m in ions what mass of na2so4 must you dissolve in125 grams of water assume total dissolution of ionic solid.
Do you mean 0.1m or 0.1M?
Do you want that total ions or that many Na and that many sulfate.
0.1 molality or m
total ions of Na2SO4
To determine the mass of Na2SO4 needed to obtain a solution with a concentration of 0.1 M in ions, we can use the formula:
moles = concentration (M) × volume (L)
First, we need to convert the given mass of water from grams to liters. The density of water is approximately 1 g/mL, so 125 grams of water is equivalent to 125 mL, which is equal to 0.125 liters.
Now, let's determine the number of moles of ions that are required in the solution. Since Na2SO4 dissociates into three ions (2 Na+ ions and 1 SO4^2- ion), the concentration of ions is three times the concentration of Na2SO4.
Concentration of ions = 3 × 0.1 M = 0.3 M
Now we can calculate the number of moles of ions using the formula mentioned earlier:
moles = concentration (M) × volume (L)
moles = 0.3 M × 0.125 L
moles = 0.0375 moles
Since one mole of Na2SO4 contains three moles of ions, we need:
moles of Na2SO4 = 0.0375 moles of ions ÷ 3
moles of Na2SO4 = 0.0125 moles
Finally, we can calculate the mass of Na2SO4 needed using its molar mass. The molar mass of Na2SO4 is:
(2 × atomic mass of Na) + atomic mass of S + (4 × atomic mass of O)
(2 × 22.99 g/mol) + 32.07 g/mol + (4 × 16.00 g/mol)
45.98 g/mol + 32.07 g/mol + 64.00 g/mol
142.05 g/mol
mass of Na2SO4 = moles of Na2SO4 × molar mass of Na2SO4
mass of Na2SO4 = 0.0125 moles × 142.05 g/mol
mass of Na2SO4 = 1.7756 grams
Therefore, approximately 1.776 grams of Na2SO4 must be dissolved in 125 grams of water to obtain a solution that is 0.1 M in ions.