﻿ Benzene and toluene form iedal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole-fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene. from Chemistry Solutions Class 12 Nagaland Board

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Benzene and toluene form iedal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole-fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Solution
Given that
PoBenzene = 50.51 mm Hg
PoNaphthalene = 50.51 mm Hg
Mass of Benzene = 80 g
Mass of Toluene = 100 g
Molar mass of benzene(C6H6) = 6 × 12  +  6 × 1  = 78 g mol - 1
Molar mass of toluene(C6H5CH3) = 6 × 12 + 5 × 1 + 12 + 3 × 1  = 92 g mol – 1
Use the formula

Mole of benzene $\left({\mathrm{n}}_{\mathrm{A}}\right)$

Mass of Toluene $\left({\mathrm{n}}_{\mathrm{B}}\right)$

Mole fraction of benzene $\left({\mathrm{x}}_{\mathrm{A}}\right)$

Similarly
Mole fraction of toluene, X Toluene =

1-XBenzene  = 1  - 0.486 = 0.514

Use the formula if Henry law
PA      = poA  ×  XA

Partial vapour pressure of benzene, PBenzene = poBenzene   ×  XBenzene

PBenzene=0.487 × 50.71  = 24.645 mm Hg
Similarly

Partial pressure of Toluene, P toluene = 0.514 × 32.06 = 16.48 mmHg

Use the formula of mole fraction using partial pressure

Plug the values we get

Mole fraction of benzene = 24.645 /(24.645 +  16.48 )

=  24.645/41.123  = 0.60

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Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,

897 Views

Calculate the molarity of each of the following solution (a) 30 g of Co(NO3)2.6H2O in 4.3 L solution (b) 30 mL of 0.5 MH2SO4 diluted to 500 mL.

solution;

Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

(a) Mol. mass of

Moles of $\mathrm{Co}\left(\mathrm{NO}{\right)}_{3}.6{\mathrm{H}}_{2}\mathrm{O}$

Volume of solution = 4.3 L
Molarity,

(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=$\frac{0.5×30}{1000}$mol =0.015mol
therefore molarity =0.015/0.5L

thus molarity is 0.03M

844 Views

Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.

(a) 20% (mass/mass) means that 20 g of KI is present in 80 g of water.

Therefore, Moles of KI in solution

moles of KI = 20/166 =0.12mol
moles of water =80/18 =4.44mol
therefore, mole fraction of KI

=

1010 Views

Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Mass % of benzene

Mass% of carbon tetrachloride = 100 - 15.28
= 84.72%
1703 Views

Calculate the mass of urea (NH2CONH2) required in making 2.5 kg 0.25 of molal aqueous solution.

Solution:

Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$
= 14 + 2 + 12 + 16 + 14 + 2
=

Molality (m) =

or Moles of solute
= 0.25 x 0.25 =  0.625

Mass of urea
= Moles of solute x Molar mass

= 0.625 x 60 = 37.5 g

1475 Views