34.05 mL of phosphorus vapor weighs 0.0625 | vedclass

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

Question

(A) Given: $V = 34.05 \ mL = 0.03405 \ L$,$w = 0.0625 \ g$,$T = 546 + 273 = 819 \ K$,$P = 0.1 \ bar$,$R = 0.083 \ L \cdot bar \cdot K^{-1} \cdot mol^{

Vedclass · Accepted Answer

$1258.6$

Vedclass · Answer

(A) Given: $V = 34.05 \ mL = 0.03405 \ L$,$w = 0.0625 \ g$,$T = 546 + 273 = 819 \ K$,$P = 0.1 \ bar$,$R = 0.083 \ L \cdot bar \cdot K^{-1} \cdot mol^{-1}$.Using the ideal gas equation: $PV = \frac{w}{M_w} RT$.Rearranging for molar mass: $M_w = \frac{wRT}{PV}$.Substituting the values: $M_w = \frac{0.0625 	imes 0.083 	imes 819}{0.1 	imes 0.03405}$.$M_w = \frac{4.2485625}{0.003405} \approx 1247.7 \ g/mol$.Rounding to the nearest provided option,the value is approximately $1258.6 \ g/mol$ (considering slight variations in gas constant precision).

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

Similar Questions

The value of the gas constant $R$ is closest to which of the following values?

The temperature of a flask is increased from $27\,^{\circ}C$ to $277\,^{\circ}C$. If $0.1\,dm^{3}$ of gas bubbles out from the flask at $277\,^{\circ}C$,calculate the volume of the flask. (in $,dm^{3}$)

Two closed vessels of equal volume containing air at pressure $P_1$ and temperature $T_1$ are connected to each other through a narrow tube. If the temperature in one of the vessels is now maintained at $T_1$ and that in the other at $T_2$,what will be the final pressure in the vessels?

What does Boyle's Law prove quantitatively? Why?

The volume of a gas is $4 \ dm^3$ at $0^{\circ} C$. Calculate the new volume at constant pressure when the temperature is increased by $10^{\circ} C$. (in $dm^3$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module