Question:
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.
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.
Show Hint
When two solutions have the same osmotic pressure, the number of moles of solute is proportional to the molar mass of the solute.
Updated On: Feb 28, 2025
Hide Solution
Verified By Collegedunia
Solution and Explanation
Osmotic pressure (\( \pi \)) is given by the formula:
\[
\pi = \frac{nRT}{V}
\]
where:
\( n \) is the number of moles of solute,
\( R \) is the gas constant,
\( T \) is the temperature,
\( V \) is the volume of the solution.
Since both solutions have the same osmotic pressure, the number of moles of solute in both solutions must be the same. For urea, the number of moles \( n_1 \) is calculated as: \[ n_1 = \frac{15 \, \text{g}}{60 \, \text{g/mol}} = 0.25 \, \text{mol}. \] Let the mass of glucose be \( m_2 \). The number of moles of glucose \( n_2 \) is: \[ n_2 = \frac{m_2}{180 \, \text{g/mol}}. \] Since the osmotic pressures are the same, the number of moles of urea must equal the number of moles of glucose: \[ 0.25 = \frac{m_2}{180}. \] Solving for \( m_2 \): \[ m_2 = 0.25 \times 180 = 45 \, \text{g}. \] Thus, the mass of glucose present in one litre of its solution is \( 45 \, \text{g} \).
\( n \) is the number of moles of solute,
\( R \) is the gas constant,
\( T \) is the temperature,
\( V \) is the volume of the solution.
Since both solutions have the same osmotic pressure, the number of moles of solute in both solutions must be the same. For urea, the number of moles \( n_1 \) is calculated as: \[ n_1 = \frac{15 \, \text{g}}{60 \, \text{g/mol}} = 0.25 \, \text{mol}. \] Let the mass of glucose be \( m_2 \). The number of moles of glucose \( n_2 \) is: \[ n_2 = \frac{m_2}{180 \, \text{g/mol}}. \] Since the osmotic pressures are the same, the number of moles of urea must equal the number of moles of glucose: \[ 0.25 = \frac{m_2}{180}. \] Solving for \( m_2 \): \[ m_2 = 0.25 \times 180 = 45 \, \text{g}. \] Thus, the mass of glucose present in one litre of its solution is \( 45 \, \text{g} \).
Was this answer helpful?
0
0
Top Questions on Solutions
- The spontaneous flow of the solvent through a semipermeable membrane from a pure solvent to a solution or from a dilute solution to a concentrated solution is called osmosis. The phenomenon of osmosis can be demonstrated by taking two eggs of the same size. In an egg, the membrane below the shell and around the egg material is semipermeable. The outer hard shell can be removed by putting the egg in dilute hydrochloric acid. After removing the hard shell, one egg is placed in distilled water and the other in a saturated salt solution. After some time, the egg placed in distilled water swells up while the egg placed in salt solution shrinks. The external pressure applied to stop the osmosis is termed as osmotic pressure (a colligative property). Reverse osmosis takes place when the applied external pressure becomes larger than the osmotic pressure.
- Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example.
- A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water \(K_f\) and \(K_b\) are 1.86 K kg mol\(^{-1}\) and 0.512 K kg mol\(^{-1}\) respectively.
- An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because __________
- Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
View More Questions
Questions Asked in CBSE CLASS XII exam
- Assertion (A): Cu cannot liberate \( H_2 \) on reaction with dilute mineral acids. Reason (R): Cu has positive electrode potential.
- CBSE CLASS XII - 2025
- Electrochemical Cells
Assertion (A): Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis. Reason (R): Aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide.
- CBSE CLASS XII - 2025
- Chemical Reactions
- Assertion (A): Vitamin D cannot be stored in our body. Reason (R): Vitamin D is fat-soluble vitamin and is not excreted from the body in urine.
- Explain \( [Co(NH_3)_6]^{3+} \) is an inner orbital complex whereas \( [Ni(NH_3)_6]^{2+} \) is an outer orbital complex. [At. No. Co = 27, Ni = 28]
- CBSE CLASS XII - 2025
- Coordination Compounds
- Calculate \( \Lambda_m^0 \) for acetic acid and its degree of dissociation (\( \alpha \)) if its molar conductivity is 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
Given that
\( \Lambda_m^0 (\text{HC}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{NaCl}) = 126 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).- CBSE CLASS XII - 2025
- Electrochemistry
View More Questions