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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(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: $1 \, bar = 750 \, torr$)

Similar Questions

Calculate the temperature of $2 \text{ moles}$ of gas occupying a volume of $5.0 \text{ L}$ at a pressure of $24.6 \text{ atm}$. (Given: $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$) (in $^\circ \text{C}$)

Explain: Gay-Lussac's Law.

$34.05 \ mL$ of phosphorus vapor weighs $0.0625 \ g$ at $546 \ ^oC$ and $0.1 \ bar$ pressure. What is the molar mass of phosphorus (in $g/mol$)?

Which amongst the following options is the correct graphical representation of Boyle's law?

Assuming ideal gas behaviour,the ratio of density of ammonia to that of hydrogen chloride at the same temperature and pressure is: (Atomic wt. of $Cl = 35.5 \ u$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module