Question

80 mole percent of MgCl₂ is dissociated in aqueous solution. The vapour pressure of 1.0 molal aqueous solution of MgCl2 at 38°C is ______ mm Hg. (Nearest Integer)

Given: Vapour pressure of water at 38°C is 50 mm Hg

Answer 1

The vapour pressure of the 1.0 molal aqueous solution of MgCl₂ at 38°C is 40 mm Hg.

Step-by-step explanation:

The vapour pressure of an aqueous solution is determined using Raoult's law. According to Raoult's law, the vapour pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapour pressure in its pure state. In this case, the vapour pressure of the aqueous solution of MgCl₂ can be calculated by multiplying the mole fraction of MgCl₂ (80 mole percent or 0.8) by the vapour pressure of pure water (50 mm Hg). Therefore, the vapour pressure of the 1.0 molal aqueous solution of MgCl₂ is 40 mm Hg (nearest integer).

Answer 2

We can use Raoult's law to calculate the vapor pressure. Vapor pressure of the solution is 40 mm Hg (Nearest Integer)

Step-by-step explanation:

To find the vapor pressure of the 1.0 molal aqueous solution of MgCl2 at 38°C, we need to calculate the mole fraction of MgCl2 in the solution. The mole fraction can be calculated using the formula:

Mole fraction of a component = Number of moles of component / Total number of moles of all components

Since 80 mole percent of MgCl2 is dissociated, the mole fraction of MgCl2 can be calculated as:

Mole fraction of MgCl2 = 0.80 * 1.0 = 0.80

The vapor pressure of the solution can be calculated using Raoult's law:

Vapor pressure of the solution = Mole fraction of MgCl2 * Vapor pressure of MgCl2

Given that the vapor pressure of water at 38°C is 50 mm Hg, we can substitute the values into the formula:

Vapor pressure of the solution = 0.80 * 50 mm Hg = 40 mm Hg (Nearest Integer)