At 27° C in a 10 L flask 4.0 g of an ideal gaseous mixture containing. He (molar mass 4.0 g mol -1 ) and Ne (molar mass 20 g mol -1 ) has a pressure of 1.23 atm. What is the mass % of neon ?(R=0.082 L atm K -1 mol -1)

Question

At 27° C in a 10 L flask 4.0 g of an ideal gaseous mixture containing. He (molar mass 4.0 g mol -1 ) and Ne (molar mass 20 g mol -1 ) has a pressure of 1.23 atm. What is the mass % of neon ?(R=0.082 L atm K -1 mol -1) (A) 25.2 (B) 62.5 (C) 84.2 (D) 74.2

Tardigrade Admin · Accepted Answer

Given T =27+273=300 K V =10 L W =W1+W2=4 g M( He ) =4 M( Ne ) =20 p =1.23 atm R =0.082 L atm K -1 mol -1 (i) ∵ pV =(n1+ n 2) RT =[(W1/4)+(W/20)] R T ∴ [(W1/4)+(W2/20)]=(p V/R T) =(1.23 × 10/0.082 × 300)=0.5 25 W1+5 W 2=50 Or, 5 W1+W2=10 ...(i) Also (ii) W1+W2=4 ...(ii) On subtracting Eq. (ii) from Eq. (i) 4 W1=6 ⇒ W1=1.5 and W2=4-1.5=2.5 Hence, Mass % of Ne (neon) =(2.5 × 100/4)=62.5 %

Q. At 27∘C in a 10L flask 4.0g of an ideal gaseous mixture containing. He (molar mass 4.0gmol−1 ) and Ne (molar mass 20gmol−1 ) has a pressure of 1.23 atm. What is the mass % of neon ?(R=0.082LatmK−1mol−1)

25.2

62.5

84.2

74.2

Q. At $2 7^{\circ} C$ in a $10 L$ flask $4.0 g$ of an ideal gaseous mixture containing. He (molar mass $4.0 g m o l^{- 1}$ ) and Ne (molar mass $20 g m o l^{- 1}$ ) has a pressure of $1.23$ atm. What is the mass $%$ of neon $? (R = 0.082 L a t m K^{- 1} m o l^{- 1})$