If 0.24 g of a volatile liquid upon vaporisation gives 48 mL of vapours at NTP. What will be the vapour density of the substance? (Density of H 2=0.089 g L -1 )

Question

If 0.24 g of a volatile liquid upon vaporisation gives 48 mL of vapours at NTP. What will be the vapour density of the substance? (Density of H 2=0.089 g L -1 ) (A) 95.39 (B) 5.993 (C) 95.93 (D) 56.0

Tardigrade Admin · Accepted Answer

Molecular mass =(0.24 × 22400/48)=112 (By Victor Meyer method) Vapour density =( text Molecular mass /2)=(112/2)=56

Q. If $0.24 g$ of a volatile liquid upon vaporisation gives $48 m L$ of vapours at NTP. What will be the vapour density of the substance? (Density of $H_{2} = 0.089 g L^{- 1}$ )

95.39

5.993

95.93

56.0

Molecular mass $= \frac{0.24 \times 22400}{48} = 112$

(By Victor Meyer method)

Vapour density $= \frac{Molecular mass}{2} = \frac{112}{2} = 56$

Q. If 0.24g of a volatile liquid upon vaporisation gives 48mL of vapours at NTP. What will be the vapour density of the substance? (Density of H2​=0.089gL−1 )