Answer:
77.58g/mol
Step-by-step explanation:
Let the rate of effusion of NO2 be R1
The rate of effusion of the unknown gas(R2) = 0.770R1
Molar Mass of NO2 (M1) = 14 + (2x16) = 14 + 32 = 46g/mol
The molar mass of the unknown gas (M2) =?
R1/R2 = √(M2/M1)
R1/0.770R1 = √(M2/46)
1/0.770 = √(M2/46)
Take the square of both sides
(1/0.770)^2 = M2/46
Cross multiply to express in linear form
M2 = (1/0.770)^2 x 46
M2 = 77.58g/mol
The molar mass of the unknown gas is 77.58g/mol