Question

estion 8 of 36 > A sample of an ideal gas has a volume of 3.70 L at 15.00 °C and 1.70 atm. What is the volume of the gas at 22.80 °C and 0.996 atm? Vs L

Answer 1

1)  

Given: P1 = 1.70 atm      V1 = 3.70 L           T1 = 15.00 ℃ ( 15 + 273 = 288 K)

            P2 = 0.996 atm     V2 = ?                   T2 = 22.80 ℃ (22.80 +273 = 295.80 K)

Apply Ideal Gas Equation,

P1V1 P2V2 T1 T2

V2 P1V1T2 T1 (P2)

V2 = 1.70 X 3.70 X 295.8 288 X 0.996

V2 = 6.48 L

2) ΔT_f = i K_f m

ΔT_f = the freezing-point depression, is defined as T_F (pure solvent) − T_F (solution).

i = van’t Hoff factor (i factor associated with the amount of dissociation of the solute in the solvent)

K_f = cryoscopic constant in ℃ kg/mol

m = molality of solute in mol solute/ kg solvent

Assuming equal concentrations, the freezing point depression will depend on van’t Hoff factor.

Al₂(SO₄)₃ – 5 ions (2Al ³⁺ and 3 SO₄^2-)

CaCl₂ – 3 ions (Ca²⁺ and 2Cl^-)

K₃PO₄ – 4 ions (3K⁺ and PO₄^3-)

NaBr – 2 ions (Na⁺ and Br^-)

Highest freezing point to lowest freezing point,

NaBr > CaCl₂ > K₃PO₄ > Al₂(SO₄)₃