Volume at NTP of 0.22 g of CO2 is same as that of

Question

Volume at NTP of 0.22 g of CO2 is same as that of (A) 0.01 g of hydrogen (B) 0.085 g of NH3  (C) 320 mg of gaseous SO2  (D) All of the above

Tardigrade Admin · Accepted Answer

Moles=(weight/molecular weight) =(volume (in L.)/22.4) ∴ (weight/molecular weight)=(molecular weight/weight× 22.4) or (1/Volume (in L))=(molecular weight/weight× 22.4) For 0.22g CO2, (1/Volume)=(44/0.22× 22.4)=(200/22.4) For 0.01 g of hydrogen, (1/Volume)=(2/0.01× 22.4)=(200/22.4) For 0.085g NH3 , (1/Volume)=(17/0.085× 22.4) =(200/22.4) For 0.32 g gaseous SO2 , (1/Volume)=(64/0.32× 22.4) =(200/22.4) Thus, all have the same volume at NTP.

Q. Volume at NTP of 0.22 g of $C O_{2}$ is same as that of

0.01 g of hydrogen

0.085 g of $N H_{3}$

320 mg of gaseous $S O_{2}$

All of the above

Q. Volume at NTP of 0.22 g of CO2​ is same as that of

0.01 g of hydrogen

0.085 g of NH3​

320 mg of gaseous SO2​

All of the above

Q. Volume at NTP of 0.22 g of $C O_{2}$ is same as that of

0.085 g of $N H_{3}$

320 mg of gaseous $S O_{2}$