Question

A clean, dried and vacuumed glass tube weighs 40.1305g. When this tube is filled with water at 240C, it comes to 138.2410 g. When filled with propylene gas, 40,2959 g comes at 240C and under 740.3 mmHg pressure conditions. What is the mole mass of propylene gas?
The same glass tube specified in the question is filled with an unknown gas at 772mmHg and 22.40C. Since the gas-filled tube is 40.4868 g, what is the molar mass of this gas?
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We have a gas mixture and the total pressure of this mixture is 2 atm. Find the mole fractions and partial pressures of each component according to the data below.
[ ] N2 > % 75,52
O2 > % 22,95
Ar > % 1,28
CO2 > % 0,25

Answer 1

a)

$$m(H_2O) = 138.2410 - 40.1305 = 98.1105g$$

$$V(H_2O) = 98.1105ml = V_{tube} = V_{gas}$$

$$m(gas) = 40.2959 - 40.1305 = 0.1620g$$

$${p_1V_1 \over T_1} = {p_2V_2 \over T_2}$$

$$V_2 = {p_1V_1T_2 \over p_2T_1}$$

$$V_2 = {760*22.4*297.15 \over 740.3*273.15} = 25L$$

$$d(gas) = m/V = 0.1620g/98.1105 mL = 0.00165g/mL$$

$$M(gas) = dV = 0.00165g/mL*25000mL/mol = 41.28g/mol$$

b)

$$m(gas) = 40.4868 - 40.1305 = 0.3563g$$

$$V_2 = {760*22.4*295.55 \over 772*273.15} = 23.86L$$

$$d(gas) = m/V = 0.3563g/98.1105mL = 0.0036g/mL$$

$$M(gas) = dV = 0.0036*23860 = 86.65g/mol$$

c)

$$p_{N_2} = 2atm*0.7552 = 1.5104atm$$

$$p_{O_2} = 2atm*0.2295 = 0.459atm$$

$$p_{Ar} = 2atm*0.0128 = 0.0256atm$$

$$p_{CO_2} = 2atm*0.0025 = 0.005atm$$

Based on the fact that volume fractions (rather than mass fractions) are given