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Chemistry I

Chemistry

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Discuss the various types of plots between the partial vapour pressure and the mole fractions of two components of the completely miscible liquids in a solution.

Answer:

When the partial vapour pressures of different (two) miscible liquids are plotted against their compositions (mole fractions), following three types of vapour pressure-composition (p – x) curves are obtained.

(a) Type-I: When the vapour pressures of the mixture lie between the vapour pressure of pure components : In such cases the solution obeys the Raoult’s law (ideal solution) i.e., the partial vapour pressure of each component (p

Fig. Solution obeying Raoult’s law.

Where x

Examples: Solution of benzene-toluene, chlorobenzene bromobenzene, hexane-heptane.

(b) Type-II : When the observed vapour pressure of the solution is greater than that of calculated vapour pressure from the Raoult’s law : In such cases partial vapour pressure of each component is found to be more than expected on the basis of the Raoult’s law. The total vapour pressure of the solution is also greater than the vapour pressure corresponding to the ideal solution. At a certain composition the total vapour pressure of the solution will be the highest (maximum) which is greater than the vapour pressure of either of the pure liquids (components) at this components the boiling point of the solution will be lowest. This type of deviation from Raoult’s law is known as positive deviation and the system exhibits a maximum value of vapour pressure at certain composition. At this composition both the liquids boil at same (constant) temperature (minimum boiling azeotropes). In figure point C, corresponds the composition of the two liquids which boils at lowest temperature. For example, alcohol-water mixture having the composition of 95.59% alcohol and 4.41 water boils at 78.13

Fig.p-x curve showing maximum in the total vapour pressure curve.

Example: Ethanol-water solution. Acetone-carbon disulphide solution. Chloroform-ethanol solution.

(c) Type-III: When the observed vapour pressure of the solution is less than that of calculated from Raoults law: In such cases the partial vapour pressure of any component is found to be less than the expected vapour pressure on the basis of Raoult’s law. Similarly, the total vapour pressure of the solution is also less than that of expected value according to Raoult’s law. At a certain composition the total vapour pressure of such solution will be lowest (minimum). At this composition the boiling point of the solution will be highest (maximum) and both the component will boil at same temperature without the change in the composition. Such composition corresponds to the maximum boiling azeotropic mixture. For example a mixture of 20.24% of HCl and 79.76% water forms an azeotropic mixture which boils at 110

Example : Solution of water-HCl, chloroform-benzene, acetone-aniline.

Fig. p-x curve showing minimum in the total vapour pressure curve.

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Calculate the mass percentage of benzene (C_{6}H_{6}) and carbon tetrachloride (CCl_{4}) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Mass % of benzene

$=\frac{\mathrm{mass}\mathrm{of}\mathrm{benzene}}{\mathrm{mass}\mathrm{of}\mathrm{solution}}\times 100\phantom{\rule{0ex}{0ex}}=\frac{22}{22+122}\times 100\phantom{\rule{0ex}{0ex}}=\frac{22}{144}\times 100=15.28\%$

Mass% of carbon tetrachloride = 100 - 15.28

= 84.72%

1703 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

=$\frac{\mathrm{moles}\mathrm{of}\mathrm{KI}}{\mathrm{moles}\mathrm{of}\mathrm{KI}+\mathrm{moles}\mathrm{of}\mathrm{water}}\phantom{\rule{0ex}{0ex}}$

$=\frac{0.12}{0.12+4.44}=0.0263$

1010 Views

Calculate the mass of urea (NH_{2}CONH_{2}) 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:

$=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}}$

Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$

= 14 + 2 + 12 + 16 + 14 + 2

= $60\mathrm{g}{\mathrm{mol}}^{-1}$

Molality (m) = $\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}}$

$25=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{2.5}$

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

Mol. mass of urea ${\mathrm{NH}}_{2}{\mathrm{CONH}}_{2}$

= 14 + 2 + 12 + 16 + 14 + 2

= $60\mathrm{g}{\mathrm{mol}}^{-1}$

Molality (m) = $\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}}$

$25=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{2.5}$

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 molarity of each of the following solution (a) 30 g of Co(NO_{3})_{2}.6H_{2}O in 4.3 L solution (b) 30 mL of 0.5 MH_{2}SO_{4} 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 $\mathrm{Co}\left({\mathrm{NO}}_{3}\right).6{\mathrm{H}}_{2}\mathrm{O}$

$=58.9+(14+3\times 16)2+6\left(18\right)\phantom{\rule{0ex}{0ex}}=58.9+(14+48)\times 2+108\phantom{\rule{0ex}{0ex}}=58.9+124+108=290.9$

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

$=\frac{30}{290.9}=0.103\mathrm{mol}.$

Volume of solution = 4.3 L

Molarity,

$\mathrm{M}=\frac{\mathrm{Moles}\mathrm{of}\mathrm{solute}}{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{in}\mathrm{litre}}\phantom{\rule{0ex}{0ex}}=\frac{103}{4.3}=0.024\mathrm{M}$

(b) Number of moles present in 1000 ml of 0.5M H_{2}SO_{4}= 0.5 mol

therefore number of moles present in 30ml of 0.5M H_{2}SO_{4}=$\frac{0.5\times 30}{1000}$mol =0.015mol

therefore molarity =0.015/0.5L

thus molarity is 0.03M

844 Views