A dry gas occupied 127.4 , mL at STP. If the | vedclass

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(A) At $STP$,$P_1 = 750 \, torr$,$V_1 = 127.4 \, mL$,and $T_1 = 273 \, K$. When collected over water,the pressure of the dry gas $(P_{dry})$ is the to

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$150$

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(A) At $STP$,$P_1 = 750 \, torr$,$V_1 = 127.4 \, mL$,and $T_1 = 273 \, K$. When collected over water,the pressure of the dry gas $(P_{dry})$ is the total pressure minus the aqueous tension: $P_2 = P_{total} - P_{H_2O} = 725 \, torr - 25 \, torr = 700 \, torr$. The temperature is $T_2 = 27 + 273 = 300 \, K$. Using the combined gas law: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$. Substituting the values: $\frac{750 	imes 127.4}{273} = \frac{700 	imes V_2}{300}$. $V_2 = \frac{750 	imes 127.4 	imes 300}{273 	imes 700} \approx 150 \, mL$.

$A$ dry gas occupied $127.4 \, mL$ at $STP$. If the same mass of the gas is collected over water at $27 \, ^\circ C$ and at a total pressure of $725 \, torr$,what volume does it occupy? The vapour pressure of water at $27 \, ^\circ C$ is $25 \, torr$. (Note: $1 \, bar = 750 \, torr$)

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