Advertisements
Advertisements
Question
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Advertisements
Solution
Given: Temperature (T) = 300 K
R = 0.083 bar L K−1 mol−1
π V = `w/(M') RT`
For the first case,
Osmotic pressure for the first case (π) = 4.98 bar
4.98 × 1 = `36/(M') xx R xx T` ...(i)
For the second case,
Osmotic pressure for the second case (π) = 1.52 bar
1.52 × 1 = `w/(M') xx R xx T` ...(ii)
Dividing eq. (i) by eq. (ii), we get,
`4.98/1.52 = 36/w`
or `w = (1.52 xx 36)/4.98`
= 10.99 g
∴ Moles of solute present in 1 L of solution = `10.99/180`
= 0.061
Hence, the molarity of the solution is 0.061 M.
RELATED QUESTIONS
Which of the following is not a colligative property?
A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution.
Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 liter of water at 25°C, assuming that it is completely dissociated.
Define Semipermeable membrane
Calculate the mass of NaCl (molar mass = 58.5 g mol−1) to be dissolved in 37.2 g of water to lower the freezing point by 2°C, assuming that NaCl undergoes complete dissociation. (Kf for water = 1.86 K kg mol−1)
Define the following term:
isotonic solution
Choose the most correct option.
In calculating osmotic pressure the concentration of solute is expressed in _______.
Answer the following in one or two sentences.
A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why?
Answer the following.
The osmotic pressure of CaCl2 and urea solutions of the same concentration at the same temperature are respectively 0.605 atm and 0.245 atm, calculate van’t Hoff factor for CaCl2.
Explain the osmotic pressure of a solution with the help of a thistle tube.
Define Osmosis.
20 g of a substance were dissolved in 500 mL of water and the osmotic pressure of the solution was found to be 600 mm of mercury at 15°C. The molecular weight of the substance is ______.
The average osmotic pressure of human blood is 7.8 bar at 37°C. What is the concentration of an aqueous NaCl solution that could be used in the blood stream?
Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/litre will be:
The temperature at which 10% aqueous solution of (W/V) of glucose will show the osmotic pressure of 16.4 atoms is: (R = 0.082 L atom K−1 mol−1)
Isotonic solutions must have the same:
(i) solute
(ii) density
(iii) elevation in boiling point
(iv) depression in freezing point
In isotonic solutions:
(i) Solute and solvent both are same.
(ii) Osmotic pressure is same.
(iii) Solute and solvent may or may not be same.
(iv) Solute is always same solvent may be different.
Match the items given in Column I and Column II.
| Column I | Column II |
| (i) Saturated solution | (a) Solution having same osmotic pressure at a given temperature as that of given solution. |
| (ii) Binary solution | (b) A solution whose osmotic pressure is less than that of another. |
| (iii) Isotonic solution | (c) Solution with two components. |
| (iv) Hypotonic solution | (d) A solution which contains maximum amount of solute that can be dissolved in a given amount of solvent at a given temperature. |
| (v) Solid solution | (e) A solution whose osmotic pressure is more than that of another. |
| (vi) Hypertonic solution | (f) A solution in solid phase. |
Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision?
Blood cells retain their normal shape in solution which are
The vapour pressure of water is 12.3 k pa at 300 k. Calculated the vapour pressure of molal solution in it.
Derive an expression to calculate molar mass of non-volatile solute by osmotic pressure measurement.
A solution containing 10 g glucose has osmotic pressure 3.84 atm. If 10 g more glucose is added to the same solution, what will be its osmotic pressure? (Temperature remains constant)
Calculate the osmotic pressure of 0.03 mole of non electrolyte solute dissolved in 0.1 dm3 of water at 300 K. [R = 0.082 dm3 atm mol-1 K-1]
The 'X' g nonvolatile solute having molar mass 196 g mol-1 is dissolved in 3 dm³ water. Calculate the value of 'X' if resulting solution has osmotic pressure 0.8 atm at 300 К.
(R = 0.0821dm³ atm K-1 mol-1)
Calculate the osmotic pressure of 0.2 M aqueous solution of nonelectrolyte at 300 K.
[R = 0.082 atm dm3 mol−1K−1]
