﻿ An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H4(OH)2) and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g ml–1, what will be the molarity of the solution? from Chemistry Solutions Class 12 Nagaland Board

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An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H4(OH)2) and 200 g of water. Calculate the molality of the solution. If the density of this solution be 1.072 g ml–1, what will be the molarity of the solution?

Molecular mass of glycol C2H4(OH)2 =62
Number of moles of C2H4(OH)2
= mass of glycol / molar mass of glycol

= 222.6/62=3.59  moles

Molality of solution =
number of moles of solute / mass of solvent in Kg

Mass of solvent = 200g = 200/1000 Kg = 0.2 kg

Plug the values we get

Molality = 3.59/0.2

Molality = 17.95

Formula of molarity of solution = number of moles of solute / volume of solution in Kg
Formula of volume = mass / density
Volume = 422.6g/(1.072 g/ml)
Volume = 394.21 ml
Convert in liter
Volume in liter = 394.21 ml /1000 liter  =0.394 liter

Molarity =9.1M
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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

=

1010 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 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 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 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