Question

What is the total mass of a mixture of 3.50x10^22 formula units na2so4, 0.500 mol h2o, and 7.23 g agcl?

Answer 1

To calculate the total mass, the masses of Na2SO4, H2O, and AgCl must be combined. The molar mass of Na2SO4 is missing, so we cannot calculate its mass. However, the mass of H2O and AgCl can be added for a partial total of 16.2375 grams.

Step-by-step explanation:

The total mass of a mixture consisting of Na2SO4, H2O, and AgCl involves calculating the mass of each component and adding them together. To find the mass of Na2SO4, we would need to use its molar mass and the number of formula units given. Unfortunately, the molar mass of Na2SO4 is not provided in the question or reference material. However, for the H2O and AgCl components, we can calculate their masses since we know the number of moles of H2O and the mass of AgCl directly.

The mass of 0.500 moles of H2O is calculated by multiplying by its molar mass (approximately 18.015 g/mol), resulting in 9.0075 grams. For AgCl, the mass is already given as 7.23 grams.

Since we cannot calculate the mass of Na2SO4 without its molar mass, we can only provide the sum of the H2O and AgCl masses in this situation, which totals to 16.2375 grams.

Answer 2

Answer is: mass of the mixture is 24,47 g.
1) N(Na₂SO₄) = 3,5·10²².
n(Na₂SO₄) = 3,5·10²² ÷ 6·10²³ 1/mol.
n(Na₂SO₄) = 0,058 mol.
m(Na₂SO₄) = 0,058 mol · 142 g/mol.
m(Na₂SO₄) = 8,24 g.
2) n(H₂O) = 0,500 mol.
m(H₂O) = 0,5 mol · 18 g/mol.
m(H₂O) = 9 g.
3) m(total) = 8,24 g + 9 g + 7,23 g.
m(total) = 24,47 g.
n - amount of substance.