The number of chlorine atoms in 20 mL of chlorine gas at STP is 1021 . (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP R =0.083 L text bar mol -1 K -1, N A =6.023 × 1023

Question

The number of chlorine atoms in 20 mL of chlorine gas at STP is 1021 . (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP R =0.083 L text bar mol -1 K -1, N A =6.023 × 1023 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

PV = nRT 1.0 × (20/1000)=( N /6.023 × 1023) × 0.083 × 273 ∴ Number of Cl 2 molecules, N =5.3 × 1020 Hence, Number of Cl-atoms =1.06 × 1021 ≈ 1 × 1021

Q. The number of chlorine atoms in $20 m L$ of chlorine gas at STP is _____ $1 0^{21}$ . (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP $R = 0.083 L bar m o l^{- 1} K^{- 1}, N_{A} = 6.023 \times 1 0^{23}$

$P V = n RT$

$1.0 \times \frac{20}{1000} = \frac{N}{6.023 \times 1 0 ^{23}} \times 0.083 \times 273$

$∴$ Number of $C l_{2}$ molecules, $N = 5.3 \times 1 0^{20}$

Hence, Number of $Cl$ -atoms $= 1.06 \times 1 0^{21}$

$\approx 1 \times 1 0^{21}$

Q. The number of chlorine atoms in 20mL of chlorine gas at STP is _____1021. (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP R=0.083L bar mol−1K−1,NA​=6.023×1023

PV=nRT 1.0×100020​=6.023×1023N​×0.083×273 ∴ Number of Cl2​ molecules, N=5.3×1020 Hence, Number of Cl-atoms =1.06×1021 ≈1×1021

Q. The number of chlorine atoms in $20 m L$ of chlorine gas at STP is _____ $1 0^{21}$ . (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP $R = 0.083 L bar m o l^{- 1} K^{- 1}, N_{A} = 6.023 \times 1 0^{23}$

$P V = n RT$

$1.0 \times \frac{20}{1000} = \frac{N}{6.023 \times 1 0 ^{23}} \times 0.083 \times 273$

$∴$ Number of $C l_{2}$ molecules, $N = 5.3 \times 1 0^{20}$

Hence, Number of $Cl$ -atoms $= 1.06 \times 1 0^{21}$

$\approx 1 \times 1 0^{21}$