0.465 g sample of an unknown compound occupies 245 mL at 298K and 1.22 atm. What is the molar mass of the unknown compound?

2 Answers
Mar 20, 2015

The molar mass of the gas is 38.0 g/mol.

One way to solve this problem is to use the Ideal Gas Law.

#PV = nRT#

#n = m/M_"r"#, where #m# is the mass and #M_r# is the relative molar mass. So,

#PV = (m/M_"r")RT#

#M_"r" = (mRT)/(PV) = ("0.465 g" × 0.082 06 cancel("L·atm·K⁻¹")"mol⁻¹" × 298 cancel("K"))/(1.22 cancel("atm") × 0.245 cancel("L")) = "38.0 g/mol"#

The relative molar mass is 38.0 g/mol.

Mar 20, 2015

Answer:

38.4g

Explanation:

#M_r=38.4#

#PV=nRT#

#P=1.22xx1.0xx10^(5)Pa#

#T=298K#

#V=245xx10^(-6)m^(3)#

#R=8.31"J/K/mol"#

#n=(PV)/(RT)=(1.22xx10^(5)xx245xx10^(-6))/(8.31xx298)#

#n=0.0121#

So 0.0121 moles weigh 0.465g

So 1 mole weighs #0.465/0.0121= 38.4"g"#

nb I converted to Pa and used 8.31 for R. It depends which R value you are given.