Question

need #7-10 answered

6. The solubility of nitrogen gas at 25°C and a nitrogen pressure of 522 mmHg is 4.7 x 10 mol/L. What is the value of the Henry's Law constant in mol/L-atm? A) 6.8 × 10-4 mol/L-atm B) 4.7 x 10 mol/L-atm C) 3.2 × 104 moVL-atm D) 9.0 x 10 mol/L atm E) 1.5 x 10mol/L atm 7. According to Raoult's law, which statement is false? A) The vapor pressure of a solvent over a solution decreases as its mole fraction increases. B) The solubility of a gas increases as the temperature decreases. C) The vapor pressure of a solvent over a solution is less than that of pure solvent. D) The greater the pressure of a gas over a solution the greater its solubility E) Ionic solutes dissociate in solution causing an enhancement of all colligative properties. 8. Calculate the freezing point of a solution made from 22.0 g of octane (CsHis) dissolved in 148.0 g of benzene, Benzene freezes at 5.50°C and its Kr value is 5.12°C/m. A)-1.16°C B)0.98°C C)6.66°C D 12.2°C549°C 9. What is the approximate Na ion concentration in a 0.75 M NazCOs solution? A) 0.375 M B 0.75 M C.25 M D) 150M E) 2.25 M 10. Calculate the molality of a 15.0% by mass solution of MgCl2 in H2O. The density of this solution is .127 g/mL. A) 0.157 m B) 11.8 m C)1.86m D) 0.0134m E) 1.58 m

Answer 1

Ans 7 : A) The vapor pressure of a solvent over a solution decreases as it mole fraction increases

As the mole fraction of a solvent increases in a solution , its vapor pressure also increases. However this vapor pressure is always less than the vapor pressure of the pure solvent.

Ans 8 :

mol octane = mass / molar mass

= 22.0 g / 114.23 g/mol = 0.192 mol

molality = mol / mass of solvent (kg)

= 0.192 / 0.148

= 1.3 m

freezing point depression = kf x m

= 5.12 x 1.3

= 6.66^oC

Freezing point of solution = freezing point of solvent - depression

= 5.50^oC - 6.66^oC

= -1.16^oC

Ans 9 :D) 1.50 M

Each mol Na2CO3 has two mol Na+ ions

So approximate concentration of Na+ ions here :

= 2 x 0.75 M

= 1.50 M

Ans 10 : C) 1.86 m

Let the mass of solution be 100 mL

So the mass of MgCl2 = ( 15 x 100) / 100 = 15 g

mass of solvent = mass of solution - mass of MgCl2

= 100 g - 15 g

= 85 g or 0.085 kg

mol MgCl2 = mass / molar mass

= 15 g / 95.211 g/mol

= 0.158 mol

Molality = mol / mass of solvent in kg

= 0.158 / 0.085

= 1.86 m