5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately

Question

5.5 mg of nitrogen gas dissolves in 180 g of water at 273 K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180 g of water at 5 atm nitrogen pressure is approximately (A) 1 × 10-5 (B) 1 × 10-4 (C) 1 × 10-6 (D) 1 × 10-3

Tardigrade Admin · Accepted Answer

Amount of nitrogen gas at 5 atm =5 × 5.5=27.5 mg =27.5 × 10-3 g Mole fraction of N 2 =( text Moles of N 2/ text Moles of H 2 O + text Moles of N 2) =( text Moles of N 2/ text Moles of H 2 O ) (∵ N 2< H 2 O ) =(27.5 × 10-3/(28)(180)18=1 × 10/-4)

Q. $5.5 m g$ of nitrogen gas dissolves in $180 g$ of water at $273 K$ and one $a t m$ pressure due to nitrogen gas. The mole fraction of nitrogen in $180 g$ of water at $5 a t m$ nitrogen pressure is approximately

$1 \times 1 0^{- 5}$

$1 \times 1 0^{- 4}$

$1 \times 1 0^{- 6}$

$1 \times 1 0^{- 3}$

Q. 5.5mg of nitrogen gas dissolves in 180g of water at 273K and one atm pressure due to nitrogen gas. The mole fraction of nitrogen in 180g of water at 5atm nitrogen pressure is approximately

1×10−5

1×10−4

1×10−6

1×10−3

Amount of nitrogen gas at 5 atm =5×5.5=27.5mg=27.5×10−3g Mole fraction of N2​= Moles of H2​O+ Moles of N2​ Moles of N2​​ = Moles of H2​O Moles of N2​​ (∵N2​<H2​O) =18180​28​27.5×10−3​=1×10−4

Q. $5.5 m g$ of nitrogen gas dissolves in $180 g$ of water at $273 K$ and one $a t m$ pressure due to nitrogen gas. The mole fraction of nitrogen in $180 g$ of water at $5 a t m$ nitrogen pressure is approximately

$1 \times 1 0^{- 5}$

$1 \times 1 0^{- 4}$

$1 \times 1 0^{- 6}$

$1 \times 1 0^{- 3}$