Question

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

• A. $\frac{16}{ N }$
• B. $\frac{32}{ N }$
• C. $16 N$
• D. $32 N$

Answer 1

Answer: (A)

$PV = nRT$

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 $=\frac{16}{ N _{ A }} g$