The volume of neon gas in cm3 at STP having the same number of atoms as that present in 800 mg of Ca is (At. mass: Ca = 40 g mol-1, Ne = 20 g mol-1)

Question

The volume of neon gas in cm3 at STP having the same number of atoms as that present in 800 mg of Ca is (At. mass: Ca = 40 g mol-1, Ne = 20 g mol-1) (A) 56 (B) 896 (C) 224 (D) 112

Tardigrade Admin · Accepted Answer

Mass of Ca =800 mg =(800/1000)=0.8 g Moles of Ca in 0.8 g=(0.8/40)=2 × 10-2 ∵ At STP, volume of 1 mole =22400 cm 3 ∴ At STP, volume of 2 × 10-2 moles =22400 × 2 × 10-2=448 cm 3

Q. The volume of neon gas in $c m^{3}$ at $STP$ having the same number of atoms as that present in $800 m g$ of $C a$ is (At. mass: $C a = 40 g m o l^{- 1}, N e = 20 g m o l^{- 1})$

56

896

224

112

448

Mass of $C a = 800 m g = \frac{800}{1000} = 0.8 g$

Moles of $C a$ in $0.8 g = \frac{0.8}{40} = 2 \times 1 0^{- 2}$

$∵$ At STP, volume of 1 mole $= 22400 c m^{3}$

$∴$ At STP, volume of $2 \times 1 0^{- 2}$ moles

$= 22400 \times 2 \times 1 0^{- 2} = 448 c m^{3}$

Q. The volume of neon gas in cm3 at STP having the same number of atoms as that present in 800mg of Ca is (At. mass: Ca=40gmol−1,Ne=20gmol−1)

56

896

224

112

448

Mass of Ca=800mg=1000800​=0.8g Moles of Ca in 0.8g=400.8​=2×10−2 ∵ At STP, volume of 1 mole =22400cm3 ∴ At STP, volume of 2×10−2 moles =22400×2×10−2=448cm3

Q. The volume of neon gas in $c m^{3}$ at $STP$ having the same number of atoms as that present in $800 m g$ of $C a$ is (At. mass: $C a = 40 g m o l^{- 1}, N e = 20 g m o l^{- 1})$

Mass of $C a = 800 m g = \frac{800}{1000} = 0.8 g$

Moles of $C a$ in $0.8 g = \frac{0.8}{40} = 2 \times 1 0^{- 2}$

$∵$ At STP, volume of 1 mole $= 22400 c m^{3}$

$∴$ At STP, volume of $2 \times 1 0^{- 2}$ moles

$= 22400 \times 2 \times 1 0^{- 2} = 448 c m^{3}$