### 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
How is osmotic pressure of a solution determined ? If the membrane used was slightly leaky, how will it influence the measured value of osmotic pressure?
Osmotic pressure of a solution containing 7 g of a protein per 103 ml of solution is 25 mm Hg at 310 K. Calculate the molecular mass or the protein. (R = 0.0821 L atm K–1 mol).

Osmotic pressure: Osmotic pressure is the minimum pressure that should be applied to the more concentrated solution to prevent osmosis.
osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus: Π = C R T
Here Π is the osmotic pressure and R is the gas constant.

(a) Measurement of Osmotic Pressure. Different methods are employed for the measurement of osmotic pressure in the laboratory but Berkley and Hartley's method gives the best results. The apparatus consists of a porous pot containing copper ferrocyanide deposited in its wall (acts as semi-permeable membrane) and fitted into a bronze cylinder to which is fitted a piston and a pressure gauge (to read the applied pressure).

The pot is fitted with a capillary indicator on left and water reservoir on right. Pot is filled with water while the cylinder is filled with a solution whose osmotic pressure is to be measured. Water tends to pass into the solution through the semipermeable membrane with the result that the water level in the indicator falls down. External pressure is now applied with piston so as to maintain a constant level in the indicator. This external pressure is osmotic pressure.
If the membrane used was a slightly, leaky, then the measured valued of osmotic pressure will not be definite.

Fig. Berkley and Hartley's apparatus.
(b)
we have given that
mB = 7g
R=0.0821 L atm K-1 mol

164 Views . 12 Shares

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 . 11 Shares

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 . 3 Shares

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 . 10 Shares

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 . 3 Shares

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 . 6 Shares

Do a good deed today
Refer a friend to Zigya