Determine the vapor pressure of a solution at 25â°c that contains 76.6 g of glucose (c6h12o6) in 250.0 ml of water. the vapor pressure of pure water at 25â°c is 23.8 torr.70.8 torr7.29 torr72.9 torr22.9 torr23.1 torr

The vapor pressure of solution is 23.1 Torr

Explanation:

Colligative property of vapour pressure.

ΔP = P° . Xm

ΔP = Pressure of vapour from pure solvent - Pressure of vapour from solution

ΔP = P° - Ps

Xm = mole fraction of solute

Molar weight of glucose: 180 g/m

Moles of glucose = mass / molar weight → 76.6 g / 180 g/m = 0.425moles

If water volume is 250 mL, this volume is occupied by 250 g of solvent.

Density of water : 1g/mL

Molar weight of water: 18 g/m

250 g / 18g/m = 13.89 moles of solvent

Total moles = moles of solute + moles of solvent

0.425 + 13.89 = 14.315 moles

Mole fraction of solute = 0.425 / 14.315 = 0.029

23.8 Torr - Ps = 23.88 Torr . 0.029

Ps = 23.88 Torr . 0.029 - 23.8 Torr ⇒ 23.1 Torr

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.