Formula units in 450 g of
is 1.93 × 10²⁴ formula units.
<u>Explanation:</u>
First we have to find the number of moles in the given mass by dividing the mass by its molar mass as,

Now, we have to multiply the number of moles of Na₂SO₄ by the Avogadro's number, 6.022 × 10²³ formula units/mol, so we will get the number of formula units present in the given mass of the compound.
3.2 mol × 6.022 × 10²³ = 1.93 × 10²⁴ formula units.
So, 1.93 × 10²⁴ formula units is present in 450g of Na₂SO₄.
The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)

Answer:
the energy possessed by a body by its value of its position relative to others, stresses within itself, electric charge, and other factors.
Explanation: