Question

Which of the following will contain maximum number of molecules
1)100L CO2 at NTP
2)10g H2 at NTP
3)1L of H20(l)
4)1000g of C12H22011(S) at NTP

Answer 1

Answer:

2)10g of H2

Explanation:

100L of CO2=100/22.4*6*10^23

10g of H2=5*6*10^23

1L of H2O=1/22.4*6*10^23

1000g of C12H22O11=1000/342*6*10^23

Max no of molecules are present in

10g H2 at NTP

Answer 2

Answer: (3) option..

Explanation:

(1) n = 100/22.4=4.46

so no of molecules = 4.46*Na

(2) n =10/2=5

so no of molecules = 5*na

(3)

1L of h20..but h20 isn't at NTP.

D=M/V

M=1000g

n=1000/18=55.5

no of molecules = 55.5*Na

(4)

n=1000/342=2.92

no of molecules = 2.92*na