Answer
There are 81.0 g of $CO_2$ in these conditions.
Work Step by Step
According to the ideal gas law:
$$PV = nRT$$
1. Choose a $R$ that has the most appropriate measurements:
Since the volume is in L, and the pressure in atm, the most appropriate R is:
$R = 0.0821 \space atm \space L \space mol^{-1} \space K^{-1}$
$T/K = 5 + 273 = 278$
2. Solving for $n$ in the ideal gas law:
$$n = \frac{PV}{RT} = \frac{(1.20 \space atm)(35.0 \space L)}{(0.0821 \space atm \space L \space mol^{-1} \space K^{-1})(278 \space K)}$$ $$n = 1.84 \space mol$$
3. The molar mass of $CO_2$ is:
$$M(CO_2) = 12.0 \space g/mol + 2*16.0 \space g/mol = 44.0 \space g/mol$$ $$m = n \times M = 1.84 \space mol \times 44.0 \space g/mol = 81.0 \space g$$
