Question

create test notebooks exam information notification profile switch to light mode logout nah! you got it wrong. jee main > chemistry > solutions start timer q₄ (numerical) mark it 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 molality of the 1.0 molal aqueous solution of MgCl₂ at 38°C is 80/95.21 g/mol, and the vapor pressure of the solution is 50 mmHg.

Step-by-step explanation:

The question pertains to the dissociation of MgCl₂ in an aqueous solution and the vapor pressure of the solution. Given that 80% of MgCl₂ is dissociated, we can determine its molality (moles of solute per kilogram of solvent). If 80 mole percent of MgCl₂ is dissociated, this means that 80% of the initial moles of MgCl₂ have dissociated into Mg²⁺ and 2Cl¯ ions. To calculate the molality, we divide the mole percent dissociated by 100 to convert it to a decimal, and then divide by the molar mass of MgCl₂. The molar mass of MgCl₂ is 95.21 g/mol. Therefore, the molality of the solution is:

molality = (80/100) / (95.21 g/mol)

The vapor pressure of the solution can be determined using Raoult's law, which states that the vapor pressure of a solution is equal to the mol fraction of the solvent multiplied by its vapor pressure at that temperature. Since the solute (MgCl₂) is nonvolatile and does not contribute to the vapor pressure, we can assume that the Mol fraction of the solvent is 1. Using the given information that the vapor pressure of water at 38°C is 50 mmHg, the vapor pressure of the solution is 50 mmHg. Therefore, the nearest integer to the vapor pressure of the 1.0 molal aqueous solution of MgCl₂ at 38°C is 50 mmHg.