A pre-weighed vessel was filled with O2 at STP and weighed. It was then evacuated, filled with SO2 at the same temperature and pressure and again weighed. The weight of O2 will be

Question

A pre-weighed vessel was filled with O2 at STP and weighed. It was then evacuated, filled with SO2 at the same temperature and pressure and again weighed. The weight of O2 will be (A) the same as that of the SO2 (B) twice as that of the SO2 (C) half of that of the SO2 (D) one-fourth of that of the SO2

Tardigrade Admin · Accepted Answer

When pressure, volume and temperature all are constant.

n_{1 (O_{2})} = n_{2 (S O_{2})}

\frac{Weight of O _{2}}{Molecular weight of O _{2}}

= \frac{Weight of S O _{2}}{Molecular weight of S O _{2}}

Weight of

O_{2}

= Weight of

S O_{2} \times \frac{32}{64}

= \frac{1}{2} \times

Weight of

S O_{2}

Q. A pre-weighed vessel was filled with $O_{2}$ at STP and weighed. It was then evacuated, filled with $S O_{2}$ at the same temperature and pressure and again weighed. The weight of $O_{2}$ will be

the same as that of the $S O_{2}$

twice as that of the $S O_{2}$

half of that of the $S O_{2}$

one-fourth of that of the $S O_{2}$

Q. A pre-weighed vessel was filled with O2​ at STP and weighed. It was then evacuated, filled with SO2​ at the same temperature and pressure and again weighed. The weight of O2​ will be

the same as that of the SO2​

twice as that of the SO2​

half of that of the SO2​

one-fourth of that of the SO2​

When pressure, volume and temperature all are constant. n1(O2​)​=n2(SO2​)​ Molecular weight ofO2​Weight ofO2​​ =Molecular weight ofSO2​Weight ofSO2​​ Weight of O2​ = Weight of SO2​×6432​ =21​× Weight of SO2​

Q. A pre-weighed vessel was filled with $O_{2}$ at STP and weighed. It was then evacuated, filled with $S O_{2}$ at the same temperature and pressure and again weighed. The weight of $O_{2}$ will be

the same as that of the $S O_{2}$

twice as that of the $S O_{2}$

half of that of the $S O_{2}$

one-fourth of that of the $S O_{2}$