### Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

### Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Vapour pressure of water
Mass of glucose W2 = 25 g
mass of water w1 = 450g
we know that
mass of glucose = 180gmol-1
molar mass of glucose =25/180g mol-1
mass of water = 18 g
molar mass of water M2 = 450/18 g mol-1
apply equation we get,

$\frac{{\mathrm{p}}_{\mathrm{A}}^{0}-{\mathrm{p}}_{\mathrm{A}}}{{\mathrm{p}}_{\mathrm{A}}^{0}}={\mathrm{x}}_{\mathrm{A}}$

$\frac{17.535-{\mathrm{p}}_{\mathrm{A}}}{17.535}=\frac{25/180}{450/18}$

or

or

or

or

or

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

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

=

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,

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%

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

Do a good deed today
Refer a friend to Zigya