Assume that helium behaves as an ideal gas.  What is the estimated density of 1.0 g of helium gas at a temperature of 27 °C and a pressure of 3.0 atm? (Note:  Use R = 0.0821 L∙atm∙mol−1∙K−1)A.1.3 × 10−4 g/mLB.5.0 × 10−4 g/mLC.1.0 × 10−3 g/mLD.1.4 × 10−3 g/mL

To solve this problem, we need to use the ideal gas law equation, which is PV = nRT.

First, we need to convert the given values to the appropriate units.

1. The temperature needs to be in Kelvin. So, we convert 27 °C to Kelvin by adding 273.15, which gives us 300.15 K.

2. The mass of helium is given as 1.0 g. We need to convert this to moles. The molar mass of helium is approximately 4 g/mol. So, 1.0 g of helium is 1.0/4 = 0.25 mol.

Now, we can substitute the given values into the ideal gas law equation:

3.0 atm * V = 0.25 mol * 0.0821 L∙atm∙mol−1∙K−1 * 300.15 K

Solving for V, we get V = (0.25 mol * 0.0821 L∙atm∙mol−1∙K−1 * 300.15 K) / 3.0 atm = 6.15 L

The density of a substance is its mass divided by its volume. So, the density of the helium gas is 1.0 g / 6.15 L = 0.1626 g/L.

However, the answer choices are given in g/mL. To convert from g/L to g/mL, we divide by 1000, so the density is 0.1626 g/L / 1000 = 1.626 x 10^-4 g/mL.

So, the closest answer choice is A. 1.3 × 10−4 g/mL.