Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
P0Pure benzene = 50.71 mm Hg
P0nepthalene = 32.06 mm Hg
Number of moles of benzene = \(\frac{39}{78}\) = 0.5 mol
Number of moles of naphthalcne = \(\frac{128}{128}\) = 1 mol
Mole fraction of benzene = \(\frac{0.5}{1.5}\) = 0.33
Mole fraction of naphthalene = 1 – 0.33 = 0.67
Partial vapour pressure of benzene =P0benzene x (Mole fraction of benzene)
= 50.71 x 0.33 = 16.73 mm Hg
Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg
Mole fraction of benzene in vapour phase = \(\frac{16.73}{16.73+21.48}\) = \(\frac{16.73}{38}\) = 0.44
Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56