Answer to Question #100294 in General Chemistry for Beverlie

Question

1. Calculate the density of H2S gas at 56 oC and 967 mmHg. Obtain the density in grams per liter.
2. A 2.30g sample of a white solid was vaporized in 345 mL vessel. If the vapor has a pressure of 985 torr at 148 oC, what is the molar mass of the solid?
3. A 3.75 L sample of gas from a chemical plant is collected and analyzed at 25oC. It is found to contain 0.0674 g NO, 0.00459 g C2H6, 0.00275 g CO2, and 0.185 g of H2O.
a. What are the partial pressures in atmospheres of each component of gas?
b. What is the total pressure in mmHg of the sample?
c. What is the mole fraction of each gas?

Answer 1

1. Use Ideal Gas Law :

$$PV = nRT$$

$$PV = \frac{m}{M}RT$$

$$\frac{m}{V} = \frac{PM}{RT}$$

$$ρ=\frac{PM}{RT}$$

T(K) = 273+56 = 329 K

p(atm) = $\frac{1 atm}{760 mm HG}\times 967 mm Hg = 1.27 atm$

$$ρ=\frac{1.27\times 34.08}{0.082\times 329} = 1.6\frac{g}{L}$$

2. Use Ideal Gas Law

$$PV = \frac{m}{M}RT$$

P(atm) =$\frac{1atm}{760 torr}\times 985 torr =1.296 atm$

T(K) 273+148=421 K

V = 0.345 L

$$1.296\times0.345 = \frac{2.30}{M}\times 0.082\times 421$$

$$M = 177.58 \frac{g}{mol}$$

3.

c.

$$n=\frac{m}{M}$$

n$(NO) = \frac{0.0674}{30.01} = 0.00225 mol$

$n(C_2H_6) = \frac{0.00459}{30.07} = 0.000153 mol$

$n(CO2) = \frac{0.00275}{44.01} = 0.0000625 mol$

$n(H2O) = \frac{0.185}{18.02} = 0.0103 mol$

$n_{tatal} = 0.00225+0.000153+0.0000625+0.0103 = 0.0128 mol$

$χ(NO) = \frac{0.00225}{0.0128} = 0.176$

$χ(C_2H_6) = \frac{0.000153}{0.0128} = 0.0120$

$χ(CO_2) = \frac{0.0000625}{0.0128} = 0.00488$

$χ(H_2O) = \frac{0.0103}{0.0128} = 0.805$

b.

$$PV=nRT$$

$$P\times 3.75 = 0.0128\times 0.082\times (273+25)$$

$$P = 0.0834 atm$$

a.

$$P_i=χ_i \times P_{total}$$

$P(NO) = 0.176\times0.0834 = 0.0147 atm$

$P(C_2H_6)= 0.0120\times 0.0834 = 0.00100 atm$

$P(CO_2) = 0.00488\times 0.0834 = 0.000407 atm$

$P(H_2O) = 0.805\times0.0834 = 0.0671 atm$