Solid iodine has a vapor pressure of 1.0 mmHg at 39 degree C. How many moles of
ID: 493259 • Letter: S
Question
Solid iodine has a vapor pressure of 1.0 mmHg at 39 degree C. How many moles of iodine will sublime into a 500 mL flask at this temperature? The vapor pressure of ethanol is 400 mmHg at 63.5 degree C. Its molar heat of vaporization is 39.3 kJ/mol. What is the vapor pressure of ethanol, in mmHg, at 34.9 degree C? Iridium has a radius of 136 pm and crystallizes in a face-centered cubic unit cell. What is the edge length of the unit cell? An element forms a body-centered cubic crystalline substance. The edge length of the unit cell is 287 pm and the density of the crystal is 7.92 g/cm^3. Calculate the atomic weight of the substance. (Avogadro's # = 6.022 times 10^23) Sodium hydroxide is available commercially as a 40.0% by weight aqueous solution. Calculate the molality of this sodium hydroxide solution. At 80 C, pure liquid A has a vapor pressure of 100 mm Hg and pure liquid B has a vapor pressure of 940 mm Hg. What is X_A for a solution of A and B with a normal boiling point of 80 C? Calculate the freezing point of a solution of 20.0 g methyl salicylate, C_7H_6O_2, dissolved in 800. g of benzene, C_6H_6. K_f is 5.10 C/m and the freezing point is 5.50 C for benzene. What is the molar mass of sucrose (table sugar) if a solution prepared by dissolving 0.822 g of sucrose in 300.0 mL of water has an osmotic pressure (pi) of 149 mm Hg at 298 K? [pi = MRT(R, gas constant = 0.08206 L.atm/K.mol)]Explanation / Answer
20.
PV = nRT, so n = PV/RT
n = (1 mm Hg)*(500 mL) / (760 mm Hg/atm)*(1000 mL/L)*(0.082 L·atm/K·mol)*(312.15 K)
= 2.57*10^-5 moles.
21.
ln(P1/P2) = Hvap/R(T1-T2/T1T2)
ln(400/P2) = 39.3/8.314(63.5-39.3 / 63.5*39.3)
= 4.73(24.2/2495.55) = 4.6475
P2 = 108.4 mmHg
22.
Using the Pythagorean Theorem
r = d / sqrt(8)
136 = d / sqrt(8)
D = 384.666 pm
23.
Let’s calculate the volume of the unit cell
V = (287 pm)^3 = (2.87 * 10^-10 m)^3
= 2.364 * 10^-29 m^3
= 2.364 * 10^-23 cm^3
Calculate the weight of the unit cell:
(2.364 * 10^-23 cm^3)*(7.92 g/cm^3) = 1.872 * 10^-22 g
There are two lattice point per bcc cell we will use this to calculate the weight per atom in the unit cell:
(1.872 * 10^-22 g)/2 = 9.61 * 10^-23 g/atom
We will now multiply the weight per atom by Avogadro's number to get the atomic weight:
(9.61 * 10^-23 g/atom)*(6.022 * 10^23 atom/mole) = 56.37 g/mole (amu)
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