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Chemistry I

Chemistry

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What is meant by positive and negative deviations from Raoult’s law and how is the sign of Δ_{mix} H related to positive and negative deviations Raoult’s law?

Answer:

According to Raoult's law, the partial vapour pressure of each component in any solution is directly proportional to its mole fraction.

The solution which obeys Raoult's law over enitre range is known as ideal solution.

The solution which do not obeys Raoult's law is known as non- ideal solution.

Non-ideal solution have vapour pressure either higher or lower is predicted by raoult's law

If the vapour pressure is higher then solution is said to exhibit positive deviation, And if the vapour pressure is lower than the solution than it said to be negtive deviation.

Positive Deviation from Raoult’s law. In those non-ideal solutions, when partial pressure of component ‘A’ in the mixture of ‘A’ and ‘B’ is found to be more than that calculated from Raoult’s law. Similarly, the partial vapour pressure of component ‘B’ can be higher than calculated from Raoult’s law.

This type of deviation from ideal behaviour is called positive deviation from Raoult’s law,

e.g., water and ethanol, chloroform and water, ethanol and CCl_{4}, methanol and chloroform, benzene and methanol, acetic acid and toluene, acetone and ethanol, methanol and H_{2}O.

For positive deviation ΔH_{mixing} > 0. (+ ve)

Negative Deviation from Raoult’s law. When the partial vapour pressure of component ‘A’ is found to be less than calculated from Raoult’s law on adding the second component ‘B’. When A is added to B, the partial vapour pressure of solution is less than that of ideal solution of same composition. Boiling point of such a solution is relatively higher than the boiling point of A and B respectively. This type of deviation from ideal behaviour is known as negative deviation from Raoult’s law e.g., chlorofom and acetone, chloroform and methyl acetate, H_{2}O and HCl, H_{2}O and HNO_{3 }acetic acid and pyridine, chloroform and benzene.

For negative deviation ΔH_{mixing} < 0.

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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,

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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

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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

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

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

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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%

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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

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