1.9 gm of a gas at 27 0C & 1.1 atm pressure occupies a volume of 1 ltr. The molecular mass of gas is, *
42.5 amu
3.83 amu
44 amu
32.5 amu

Respuesta :

Answer:

42.5 amu

Explanation:

Using the ideal gas law equation;

PV = nRT

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information provided in this question;

mass = 1.9g

T = 27°C = 27 + 273 = 300K

P = 1.1atm

V = 1L

n = ?

PV = nRT

1.1 × 1 = n × 0.0821 × 300

1.1 = 24.63n

n = 1.1/24.63

n = 0.045mol

Mole = mass/molecular mass

0.045 = 1.9/M.M

M.M = 1.9/0.045

M.M = 42.54amu