﻿ Derive a relationship between mole fraction and vapour pressure of a compound of an ideal solution in the liquid phase and vapour phase. from Chemistry Solutions Class 12 Nagaland Board

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Derive a relationship between mole fraction and vapour pressure of a compound of an ideal solution in the liquid phase and vapour phase.

for any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.
pA ∝ xA

Where only solvent is volatile
pA a xA where p A is vapour pressure of solvent having mole fractionxA,
But xA + xB = 1
xA = 1 – xB where xB is mole fraction of non-volatile solute B
pA = p0A (1 – xB)
= p0A – p0A x B
Total vapour pressure

Solution containing non-volatile solute : For a solution of non-volatile solid in a liquid the vapour pressure contribution by the non-volatile solute is negligible. Therefore the partial vapour pressure of a solution containing a non-volatile solute is equal to the product of the vapour pressure of the pure liquid (solvent p0A) and its mole fraction in solution.
PA = P0A x xB ....(i)
xB is the mole fraction of the non-volatile solute
B, then xA + xB = 1
xA = 1 – xB ....(ii)
Substituting the value of xA fromeq. (ii) into eq. (i), we get, pA = p0A (1 – xB) = p0A – p0x B
$\frac{{{\mathrm{p}}^{0}}_{\mathrm{A}}-{\mathrm{p}}_{\mathrm{A}}}{{{\mathrm{p}}^{0}}_{\mathrm{A}}}={\mathrm{x}}_{\mathrm{B}}.$

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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 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 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

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