Calculate the mass of ethylene glycol (c₂h₆o₂) that must be added to 1.00 kg of ethanol (c₂h₅oh) to reduce its vapor pressure by 11.0 torr at 35°c. the vapor pressure of pure ethanol at 35°c is 1.00×10² torr.

123.5 g of C₂H₅OH are required

Explanation:

Let's think the colligative property of vapour pressure to solve this.

ΔP = P°. Xm

ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

P° = Vapor pressure of pure solvent

Xm = mole fraction of solute (moles of solute / (moles of solute + moles of solvent))

11 Torr = 100 Torr . Xm

11 Torr / 100 Torr = 0.11 → Xm

0.11 = moles of solute / moles of solute + moles of solvent

Let's determine the moles of solvent.

We have a mass of 1 kg, which is the same as 1000 g

Molar mass ethanol = 46 g/mol

1000g / 46g/mol = 21.7 moles

0.11 moles = moles of solute / moles of solute +21.7 moles

0.11 moles (moles of solute +21.7 moles) = moles of solute

0.11 moles of solute + 2.39 moles = moles of solute

2.39 moles = 1 - 0.11 moles of solute

2.39 moles = 0.89 moles of solute

2.39 / 0.89 = moles of solute → 2.68 moles

Let's convert the moles to mass

2.68 moles . 46 g/mol = 123.5 g

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.