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
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
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