1) CaCl is a strong electrolyte which dissociates in aqueous solution according

Question

1) CaCl is a strong electrolyte which dissociates in aqueous solution according to the following: CaCl2 Ca2+ + 2 cr 0.384 moles of CaCl2 is dissolved in 300 g of water at 180C. The vapor pressure of pure water at 18°C is 0.0204 atm. a. Compute the number of moles of solute particles present. c. Compute the vapor pressure of water over this solution. b. Compute the mole fraction of solute particles and the mole fraction of water. 2) Compute the number of moles of solute particles in each of the following: 0.34 moles of KBr a. b. 0.98 g of CH3OH (a non-electrolyte) c. 13.8 g of NaF d. 64.7 g of FeCls Compute the molality of the solution, Tb , and the boiling point for each of the following: a. b. c. d. 0.24 moles of CH3OH (a non-electrolyte) dissolved in 160 g of water 0.75 moles of CaClz dissolved in 13.8 moles of water 125.0 g of a solution which contains 15.8 g of Ca(OH)2 de-aeuhwatn 10.5 g of FeCls dissolved in 140 g of water

Explanation / Answer

1) a) According to the given equation, 1 mole of CaCl2 dissociates in aqueous solution into 1 mole of Ca2+ ions and 2 moles of Cl- ions, i.e. a total of 3 moles of ions or solute particles.

Here, the no. of moles of CaCl2 = 0.384 mol

Therefore, the no. of moles of solute particles present (nsolute) = 3*0.384 = 1.152 mol

b) The molar mass of water = 18 g/mol

The mass of water = 300 g

No. of moles = mass/molar mass

i.e. The no. of moles of water (nwater) = 300 g/18 g mol-1 = 16.667 mol

The total no. of moles (n) = nsolute + nwater = 1.152 + 16.667 = 17.819 mol

The molefraction of solute particles (xsoltue) = nsolute/n

i.e. xsoltue = 1.152/17.819 = 0.065

The molefraction of water (xwater) = 1-0.065 = 0.935

c) According to Roult's law of vapor pressures: Pwater = xwater * Powater

Where Pwater = vapor pressure of water over the given solution, Powater = vapor pressure of pure water

i.e. Pwater = 0.935*0.0204 atm = 0.0191 atm

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