A `19.6 g` of a given gaseous sample contains `2.8 g` of molecules `(d = 0.75 g L^(-1))`, `11.2 g` of molecules `(d = 3 g L^(-1))` and `5.6 g` of molecules `(d = 1.5 g L^(-1))`. All density measurements are made at `STP`. Calculate the total number of molecules `(N)` present in the given sample. Report your answer in `10^(23) N`
Assume Avogardro's number as `6 xx 10^(23)`.

Answer

C

Explanation

Total weight `= 19.6 g`
a. `2.8 g` of molecules, `d = 0.765 g L^(-1)`
volume `= ("Mass")/(d) = (2.08)/(0.75) L` at `STP`
1 mol `-= 22.4 L` at `STP`
moles `= (2.8)/(0.75) xx (1)/(22.4) xx 6 xx 10^(23) = 1 xx 10^(23)`
b. `11.2 g` molecules, `d = 3 g L^(-1)`
Molecules `= 1 xx 10^(23)`
c. `5.06 g` molecules, `d = 1.5 g L^(-1)`
Molecules `= 1 xx 10^(23)`
Total number of molecules `= 3 xx 10^(23) = 3`