The weight of 350 mL of a diatomic gas at 0° C and 2 atm pressures is 1 g. The weight of one atom in gram is: (N = Avogadro's number)

Question

The weight of 350 mL of a diatomic gas at 0° C and 2 atm pressures is 1 g. The weight of one atom in gram is: (N = Avogadro's number) (A) (16/ N ) (B) (32/ N ) (C) 16 N (D) 32 N

Tardigrade Admin · Accepted Answer

PV = nRT Let the atomic weight be A 2 × (350/1000)=(1/2 A ) × 0.0821 × 273 A=(224/14)=16 Mass of one atom =(16/ N A ) g

Q. The weight of $350 m L$ of a diatomic gas at $0^{\circ} C$ and $2 a t m$ pressures is $1 g$ . The weight of one atom in gram is: (N = Avogadro's number)

$\frac{16}{N}$

$\frac{32}{N}$

$16 N$

$32 N$

$P V = n RT$

Let the atomic weight be A

$2 \times \frac{350}{1000} = \frac{1}{2 A} \times 0.0821 \times 273$

$A = \frac{224}{14} = 16$

Mass of one atom $= \frac{16}{N _{A}} g$

Q. The weight of 350mL of a diatomic gas at 0∘C and 2atm pressures is 1g. The weight of one atom in gram is: (N = Avogadro's number)

N16​

N32​

16N

32N