﻿ Vapour pressures of pure acetone and chloroform at 328 k are 741.8 nm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal , Pchlroform and Pacelone as a function of Xactone. The experimental date observed for different composition of mixture is: 100 x xacetone 0 11.8 23.8 36.0 50.8 58.2 64.5 72.1 Pacetone / mm Hg   0 54.9 110.1 202.4 327.7 405.9 454.1 521.1 Pchloroform/ mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7 Plot this data also on the same graph paper, indicate whether it has positive deviation or negative deviation from the ideal solution. from Chemistry Solutions Class 12 Nagaland Board

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Vapour pressures of pure acetone and chloroform at 328 k are 741.8 nm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot Ptotal , Pchlroform and Pacelone as a function of Xactone. The experimental date observed for different composition of mixture is:

 100 x xacetone 0 11.8 23.8 36 50.8 58.2 64.5 72.1 Pacetone / mm Hg 0 54.9 110.1 202.4 327.7 405.9 454.1 521.1 Pchloroform/ mm Hg 632.8 548.1 469.4 359.7 257.7 193.6 161.2 120.7

Plot this data also on the same graph paper, indicate whether it has positive deviation or negative deviation from the ideal solution.

The plot is given below:

It has negative deviation from the ideal solution.

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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 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