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Answer:
4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Explanation:
By ideal gas equation:

Number of moles (n)
can be written as: 
where, m = given mass
M = molar mass

where,
which is known as density of the gas
The relation becomes:
.....(1)
We are given:
M = molar mass of chloroform= 119.5 g/mol
R = Gas constant = 
T = temperature of the gas = 
P = pressure of the gas = 1.00 atm
Putting values in equation 1, we get:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.
Answer:
we know, at STP ( standard temperature and pressure).
we know, volume of 1 mole of gas = 22.4L
weight of 1 Litre of hydrogen gas = 0.09g
so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.
similarly,
weight of 2L of a gas = 2.88gm
so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g
hence, molecular weight of the gas = 32.256g
vapor density = molecular weight/2
= 32.256/2 = 16.128g
hence, vapor density of the gas is 16.128g.
Explanation:
Covalent since they are both non metals, a metal and a non metal make an ionic bond