﻿ The molar mass of acetic acid determined from the measurement of a colligative property is greater than its formula molar mass. Explain why? from Chemistry Solutions Class 12 Nagaland Board

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The molar mass of acetic acid determined from the measurement of a colligative property is greater than its formula molar mass. Explain why?

The colligative properties of solution depend on the total number of solution.
since the eletrolytes ionise and give more than one particle per formula unit in solution the colligative effect of an eletrolyte solution is alaways greater than that of a non electrolyte of the same molar concentration.

The molar mass of acetic acid in a solution of benzene solvent is greater than its formula molar mass because two CH3COOH molecules form a dimer (CH3COOH)2 in solution. But we know that the molar mass is defined as the mass of 6.023 x 1023 particles of the substance. In case of a normal solute like CH3COOH, the molar mass of acetic acid is equal to the mass of 6.023 x 1023 molecules of formula CH3COOH. While in solution the solution dimerises and becomes a bigger molecule like (CH3COOH)2. Now in this case the molar mass of the solute is equal to the mass of 6.023 x 1023 units of formula (CH3COOH)2. Naturally, this molar mass is larger than the normal formula molar mass. In case of association, Van’t Hoff factor is less than 1, therefore,

MB (abnormal) > MB (normal) when i < 1.

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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 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 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