1.

A person accidentally swallow a drop of liquid oxygen, O_(2)(l) which as density 1.2 gm/ml. Assuming drop has volume 0.05 ml. What volume of gas will be produced in person's stomach at a body temperature (27^(@)C) and pressure 1 atm. [Take R=0.08 atm-lit K^(-1) "mol"^(-1)]]

Answer»

40ml
30ml
50ml
45ml

Solution :`d=(m)/(V)rArrm_(O_(2)(L))=1.2xx0.005=0.060gm`
`n_(O_(2))=(0.06)/(32)`
`n_(O_(2))=(NRT)/(P)=(6xx0.08)/(32xx100)xx300`
V=45ml


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