To get the empirical formula of this compound, we take a basis of 100 grams which means each percentage is equivalent to 1 gram. Hence there is 32.39 grams sodium, 22. 53 grams sulfur and 45.07 grams oxygen. We convert each mass to their moles by dividing by their respective molar mass. Na: 1.408, S:0.704 and O:2.82. divide each with the lowest: Na: 2: S: 1 and O:4. Hence the formula is Na2SO4.
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Answer: -
3.151 M
Explanation: -
Let the volume of the solution be 1000 mL.
At 25.0 °C, Density = 1.260 g/ mL
Mass of the solution = Density x volume
= 1.260 g / mL x 1000 mL
= 1260 g
At 25.0 °C, the molarity = 3.179 M
Number of moles present per 1000 mL = 3.179 mol
Strength of the solution in g / mol
= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)
Now at 50.0 °C
The density is 1.249 g/ mL
Mass of the solution = density x volume = 1.249 g / mL x 1000 mL
= 1249 g.
Number of moles present in 1249 g = Mass of the solution / Strength in g /mol
= 
= 3.151 moles.
So 3.151 moles is present in 1000 mL at 50.0 °C
Molarity at 50.0 °C = 3.151 M