Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL -1 at 257° C and 100 mm Hg. The molar mass of geraniol is (Nearest Integer) [Given R =0.082 L atm K -1 mol -1 ]

Question

Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46 gL -1 at 257° C and 100 mm Hg. The molar mass of geraniol is (Nearest Integer) [Given R =0.082 L atm K -1 mol -1 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Assuming ideal behaviour P =( dRT / M ) P =(100/760) atm , T =257+273=530 K d =0.46 gm / L So M =(0.46 × 0.082 × 530/100) × 760 =151.93 ≈ 152

Q. Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is $0.46 g L^{- 1}$ at $25 7^{\circ} C$ and $100 mm H g$ . The molar mass of geraniol is _____(Nearest Integer) [Given $R = 0.082 L a t m K^{- 1} m o l^{- 1}$ ]

Assuming ideal behaviour $P = \frac{d RT}{M}$
$P = \frac{100}{760} a t m, T = 257 + 273 = 530 K$
$d = 0.46 g m / L$
So $M = \frac{0.46 \times 0.082 \times 530}{100} \times 760$
$= 151.93 \approx 152$

Q. Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is 0.46gL−1 at 257∘C and 100mmHg. The molar mass of geraniol is _____(Nearest Integer) [Given R=0.082LatmK−1mol−1 ]

Assuming ideal behaviour P=MdRT​ P=760100​atm,T=257+273=530K d=0.46gm/L So M=1000.46×0.082×530​×760 =151.93≈152

Q. Geraniol, a volatile organic compound, is a component of rose oil. The density of the vapour is $0.46 g L^{- 1}$ at $25 7^{\circ} C$ and $100 mm H g$ . The molar mass of geraniol is _____(Nearest Integer) [Given $R = 0.082 L a t m K^{- 1} m o l^{- 1}$ ]

Assuming ideal behaviour $P = \frac{d RT}{M}$
$P = \frac{100}{760} a t m, T = 257 + 273 = 530 K$
$d = 0.46 g m / L$
So $M = \frac{0.46 \times 0.082 \times 530}{100} \times 760$
$= 151.93 \approx 152$