At a temperature of 18^{circ}C and a pressure | vedclass

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

Question

(A) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,the molar mass $M$ is given by $M = \frac{mRT}{PV}$.Given: $m = 2.71 \, g$,$P = \frac{765}

Vedclass · Accepted Answer

$49.5$

Vedclass · Answer

(A) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,the molar mass $M$ is given by $M = \frac{mRT}{PV}$.Given: $m = 2.71 \, g$,$P = \frac{765}{760} \, atm$,$V = 1.299 \, L$,$T = 18 + 273.15 = 291.15 \, K$,and $R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$.Substituting the values: $M = \frac{2.71 	imes 0.0821 	imes 291.15}{(765/760) 	imes 1.299}$.$M = \frac{64.76}{1.307} \approx 49.5 \, g/mol$.

At a temperature of $18^{\circ}C$ and a pressure of $765 \, torr$,the weight of $1.299 \, L$ of a gas is $2.71 \, g$. Calculate the molar mass of the gas at $STP$.

Similar Questions

Which of the following expressions at constant pressure represents Charles's law?

By what other name is the ideal gas equation known? Why?

For a fixed mass of an ideal gas,the following graph is given. Determine the correct relationship between $P_1, P_2$ and $P_3$.

One mole of an ideal gas at $300 \ K$ is expanded isothermally from $1 \ L$ to $10 \ L$ volume. $\Delta U$ for this process is $....... \ J$ (Use $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

At very low pressure,the compressibility factor $Z$ of a $CO_2$ gas with constant molar volume is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module