Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. a. 0.100 m K₂S b. 21.5 g of CuCl₂ in 4.50⨉10² g water

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Hi everyone here, we have a question asking us to determine the boiling point and freezing point of a solution that is prepared by dissolving 13.5 g of nickel bromide in 10. times 10 to the second grams of water. Assume that nickel bromide completely dissociates in water. So we're going to calculate the morality first. The morality equals the moles of solute, divided by the kilograms of solvent. Our molar mass of nickel bromide equals 0.69 grams per mole plus .904 grams per mole times two. And that equals .498 g per mole. Our models of nickel bromide equals 13 0. nickel bromide times one mole nickel bromide divided by its smaller mass. So .498 g of nickel bromine. And our grams here are going to cancel out. And that equals 0. moles of nickel bromide. So our mass of water Equals 2.60 Times 10 to the 2nd g of water times one kg over to the 3rd g. So our g are canceling out here, leaving us with zero .260 kg of water. Our mobility equals 0.0 moles of nickel bromide Divided by 0. kg. And that equals 0.2376 morality. So now we are assuming that nickel bromide completely dissolves into nickel plus two bromine and that makes it an electrolyte with event half factor of three because we have one nickel and two bromide, Our change in our boiling point is going to equal boiling point of our solution minus the boiling point of our solvent. So are boiling point is going to equal our van halt factor times are boiling point constant times our morality, Our boiling point constant equals zero .512° C per morality. Our boiling point of our solvent equals 100°C. So our change in boiling point Is going to equal three Times 0.512 degrees Celsius per morality times 0.2376 Morality And that equals 0.365°C. So our boiling point of our solution is going to equal a change boiling point plus our boiling point of our pure solvent, which is 100 degrees Celsius plus 0.365 degrees Celsius, Which equals .365°C. And that is our first answer. Next. We're going to solve for the freezing point. So the change in our freezing point equals the freezing point of our pure solvent minus the freezing point of our solution. Our change in freezing point is going to equal our van Hoff factor times our freezing point constant times morality, our Freezing point constant equals 1.86 degrees Celsius per morality. And the freezing point of our peer solvent equals 0°C. So our change in freezing point equals three Times 1.86 degrees Celsius per morality Times 0. Morality, And that equals 1.326°C. Our freezing point of our solution equals are freezing point of our pure solvent minus our freeze, our change in freezing point, So that equals 0°C -1.326°C Which equals negative 1.326°C which is our second and final answer. Thank you for watching. Bye.

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. a. 0.100 m K₂S b. 21.5 g of CuCl₂ in 4.50⨉10² g water

Verified Solution

Key Concepts

Colligative Properties

Dissociation of Ionic Compounds

Freezing Point Depression and Boiling Point Elevation Formulas

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. a. 0.100 m K2S b. 21.5 g of CuCl2 in 4.50⨉102 g water

Verified Solution

Key Concepts

Colligative Properties

Dissociation of Ionic Compounds

Freezing Point Depression and Boiling Point Elevation Formulas

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. a. 0.100 m K₂S b. 21.5 g of CuCl₂ in 4.50⨉10² g water