An unknown gas occupies a volume of 4.75 l at 1227 "cand 5.00 atm. if the mass is 5.45 g what is the molar mass of the gas? (r=0.0821 atml/molk) a) 23.8 g/mol b) 344 g/mol c) 215 g/mol d) 28.3 g/mol e) 141 g/mol

Answer : The correct option is, (D) 28.3 g/mole

Explanation :

Using ideal gas equation :

where,

P = pressure of gas = 5.00 atm

V = volume of gas = 4.75 L

T = temperature of gas =

n = number of moles of gas

w = mass of gas = 5.45 g

M = molar mass of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

Therefore, the molar mass of the gas is 28.3 g/mole

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :