An aqueous solution containing 35.9 g of an unknown molecular (nonelectrolyte) compound in 150.0 g of water has a freezing point of —1.3 °C. Calculate the molar mass of the unknown compound.

molr mass (mm):

mm = 342.43 g/mol

Explanation:

freezing point (ΔTf):

∴ ΔTf = - 1.3°C

∴ a: water (solvent)

∴ b: solute

∴ mb: molality of the solute [=] mol/Kg

∴ Kf: cryoscopic constant

⇒ Kf water = 1.86 °C.Kg/mol

∴ wb = 35.9 g

∴ wa = 150.0 g = 0.150 Kg

∴ molar mass solute (mm) [=] g/mol

⇒ mb = - ΔTf/Kf

⇒ mb = - ( - 1.3°C)/(1.86 °CKg/mol)

⇒ mb = 0.6989 mol/Kg

∴ moles solute (nb):

⇒ nb = (0.6989 mol/Kg)(0.150 Kg) = 0.1048 mol

molar mass:

⇒ mm = (35.9 g)/(0.1048 mol) = 342.43 g/mol

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

it is the most commonly accepted one