Question

Q. Cal volume occupied by A equimolar mixture of N2 & O2 which
weighs 30gm at NTP.​

Answer 1

mass =30g

sum of molar mass=60g/mol

noof moles =mass/molar mass

30/60=0.5mol

total n of moles is 0.5mol

now the ideal gas equation is

pv=nrt

at NTP t is 273k p is 1atm

v=nrt/p

v=0.5×273×0.0821/1=11.2L

r is constant 0.0821 l atm /k mol

11.2L is the answer