Answer:
Molarity of the solution = 21.6 M
Explanation:
<u>Given:</u>
Moles of CH#COOH = 4.32*10^2 moles
Volume of water = 20.0 L
<u>To determine:</u>
Molarity of the solution
<u>Explanation:</u>
Concentration of a given solution can be expressed in terms of Molarity which refers to the moles of solute per liter volume of the solution
For the given CH3COOH solution:
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.
In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.
Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.
To calculate molecular formula of compound, convert mass into number of moles as follows:
Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.
Taking the ratio:
Putting the values,
Thus, molecular formula of compound will be .
