How many molecules are there in 79g of Fe2O3? how many atoms is this?

A mixture is created by dissolving 1.43 mol of potassium chloride (KCl) in 889 g of water. The concentration of KCl works out to be 1.61 molal.
The amount of KCl is 1.43 mol
Water weighs 889 g
The molality can be calculated using the formula:
molality = moles of solute divided by kilograms of solvent
Since 1 kg equals 1000 g, 889 g is 0.889 kg.

Therefore, m = 1.43/0.889 = 1.61 molal.

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Identifying the given parameters from the question;

Vapor pressure = 94.4 mm of Hg

The reaction for vaporization is expressed as;

C₆H₆(l) ⇄ C₆H₆(g)

The equilibrium in terms of activities can be defined as:

K = a(C₆H₆(g)) / a(C₆H₆(l))

The activity for pure substances equals one:

a(C₆H₆(l)) = 1

For an ideal gas phase, activity is approximated as the ratio of partial pressure to total pressure. Under standard conditions:

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg is the standard pressure

Thus, we find;

K = 94mmHg / 760mmHg = 0.12421

The formula for Gibbs free energy is:

ΔG = - R·T·ln(K)

Here, R represents the gas constant = 8.314472J/molK

Consequently, the ΔG° for the vaporization of benzene is calculated as:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol

The change in Gibbs free energy for the reaction is determined by the difference between the Gibbs free energy of formation of the products and reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

<pThus:

ΔfG°(C₆H₆(g)) = ΔvG° + ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol