Answer
(a) $\rho = 1.34~kg/m^3$
(b) $\rho = 1.16~kg/m^3$
Work Step by Step
(a) We can convert the average molecular mass to units of $kg$:
$m = (29~u)(1.66\times 10^{-27}~kg/u) = 4.814\times 10^{-26}~kg$
We can find an expression for the number density:
$P~V = N~k~T$
$\frac{N}{V} = \frac{P}{k~T}$
We can find an expression for the mass density:
$\rho = \frac{m~N}{V}$
$\rho = \frac{m~P}{k~T}$
We can find the mass density when the temperature is $-10^{\circ}C$ which is $263~K$:
$\rho = \frac{m~P}{k~T}$
$\rho = \frac{(4.814\times 10^{-26}~kg)(1.01\times 10^5~Pa)}{(1.38\times 10^{-23}~J/K)(263~K)}$
$\rho = 1.34~kg/m^3$
(b) We can find the mass density when the temperature is $30^{\circ}C$ which is $303~K$:
$\rho = \frac{m~P}{k~T}$
$\rho = \frac{(4.814\times 10^{-26}~kg)(1.01\times 10^5~Pa)}{(1.38\times 10^{-23}~J/K)(303~K)}$
$\rho = 1.16~kg/m^3$
