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BartSMP [9]
3 years ago
15

A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc

ed the following data: Volume of O2 produced at room conditions 280 mL Barometric pressure 740 torr Temperature of water 24°C Termperature of O2 25°C Vapor Pressure due to water at 25°C 22.4 torr For the conditions listed above, calculate the volume of O2(g) produced at standard conditions of temperature and pressure. (enter your answer in liters)
Chemistry
1 answer:
Paul [167]3 years ago
5 0

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

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Given mass of diethyl ether = 208.0 g

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Putting values in equation 1, we get:

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\chi_A=\frac{n_A}{n_A+n_B}

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\chi_{\text{solute}} = mole fraction of solute (testosterone) = 0.0095

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\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg

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