Question #75905
1 Answer
Sodium.
Explanation:
The idea here is that you need to use the ideal gas law equation to find a relationship between the density of the gas and its molar mass.
As you know, the ideal gas law equation establishes a relationship between pressure and volume, on one side, and number of moles and temperature on the other.
#PV = nRT#
Here
#R = 0.082("atm" * "L")/("mol" * "K")#
Now, the number of moles of can be written as the ration between the mass of the sample and the gas' molar mass
#n = m/M_M#
Plug this into the idea lgas law equation to g et
#PV = m/M_M * RT#
Multiply both sides of the equation by
#PV * M_M = m/color(red)(cancel(color(black)(M_M))) * color(red)(cancel(color(black)(M_M))) * RT#
#PV * M_M = m * RT#
Now look what happens when you divide both sides by the volume of the gas
#(Pcolor(red)(cancel(color(black)(V))) M_M)/color(red)(cancel(color(black)(V))) = m/V * RT#
#P * M_M = m/V * RT#
But since density is defined as mass per unit of volume, you will have
#P * M_M = rho * RT#
Now plug in your values and solve this equation for
#M_M = (rho * RT)/P#
#M_M = (2.9 * 10^(-3)"g"/color(red)(cancel(color(black)("L"))) * 0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (1000 + 273.15)color(red)(cancel(color(black)("K"))))/((10.)/760color(red)(cancel(color(black)("atm"))))#
#M_M = "23.009 g/mol" ~~ "23 g/mol"#
Now, the important thing to realize here is that you're dealing with a gaseous element, not a compound.
The high temperature at which the element is kept is a clue - in this case, sodium has a molar mass of approximately
This means that at