A solution contains 6 mg isopropanol dissolved in 2 kg H2O. The isopropanol concentration is a.3% isopropanol.
b.3 M isopropanol.
c.3 m isopropanol.
d.3 ppm isopropanol.

Option D. 3 ppm isopropanol

Explanation:

The following data were obtained from the question:

Mass of isopropanol = 6 mg

Mass of water = 2 Kg

Isopropanol concentration =?

The isopropanol concentration can be obtained as follow:

Concentration = mass isopropanol / mass of water

Isopropanol concentration = 6 mg / 2 kg

Isopropanol concentration = 3 mg/Kg

Finally, we shall convert 3 mg/Kg to ppm. This can be obtained as follow:

1 mg/Kg = 1 ppm

Therefore,

3 mg/Kg = 3 ppm

Thus, we can conclude that the concentration of isopropanol is 3 ppm.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

2b2t

explanation:

2b2t