The solubility of a gas in general _____ with increasing pressure and _____ with

Question

The solubility of a gas in general _____ with increasing pressure and _____ with increasing temperature. (A) decreases, increases (B) decreases, decreases (C) increases, increases (D) increases, decreases The molality of a solution of ethyl alcohol, C_2H_5OH, in water is 0.96 mol/kg. How many grams of alcohol are dissolved in 5.98 kg of water? (A) 287.2 g (B) 7.40 g (C) 74.1 g (D) 264.7 g What is the mass of C_12H_22O_11 in 60.0 mL of 0.0880 M solution? [MM C_12H_22O_11-342.3 g/mol] (A) 0.181 g (B) 5.28 g (C) 1.81 g (D) 5.02 g What is the molarity of a solution that is 46% by mass perchloric acid (HClO_4)? The density of the solution is 1.23 g/mL.[MM HClO_4 - 100.46 g/mol] (A) 7.9 M (B) 1.2 M (C) 5.6 M D) none of the above What is the percent KCl by mass in a 3.00 molal solution? (MM KCl - 74.55 g/mol] (A) 30.0% (B) 18.3% (C) 0.300% (D) 12.5%

Explanation / Answer

1. Ans D

For Gases, solubility decreases as temperature increases. When most gases dissolve in solution, the process is exothermic. This means that heat is released as the gas dissolves. As temperature increases, kinetic energy increases. The higher kinetic energy causes more motion in the gas molecules which break intermolecular bonds and escape from solution. Gases increase in solubility with an increase in pressure. Henry's Law states that: The solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.

2. molality = no. of moles of solute in 1Kg of solvent =0.96 mol/Kg

No of moles of ethanol in 5.98kg water = 0.96 mol/Kg x 5.98 Kg = 5.74 mol

Mass /molecular mass = 5.74 mol

Mass / 46.07 g/mol = 5.74 mol

Mass = 264.7 g

3. molarity = no. of mols of solute in 1L of solvent = 0.088 M

No. of moles in 60 ml solution = 0.088 mol/Lx 0.060 L = 0.00528 mol

molar mass of C12H22O11 = 342.3 g/mol

mass/molarmass =0.00528 mol

mass = 0.00528 mol x 342.3 g/mol = 1.81 g

4. 46 % HClO4

Density of HClO4 = 1.23 g/mL

Mass of 1L solution = 1000 mL x 1. 23g/mL = 1230 g

Mass of HClO4 = 1230 x 46/100 = 565.80 g

Moles of HClO4 = 565.80 / 100.46 g/mol = 5.63

Molarity = no. of moles in 1L solution = 5.6 M

ans is c

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