Question

A $10\% $ of cane sugar (molecular weight $342$) is isotonic with $1.754\% $ aqueous solution of urea. Find the molecular mass of urea.

Answer 1

Hint: The solutions having the same concentration and osmotic pressure at a given temperature is termed as isotonic solutions. Thus we can say that at a given temperature, osmotic pressure is directly proportional to the concentration of the solution.

Complete step by step answer:
It is given that the weight by volume percentage of cane sugar, $\dfrac{{{w}}}{{{v}}}{\% _{{{sug}}}} = 10\% $
Weight by volume percentage of urea, $\dfrac{{{w}}}{{{v}}}{\% _{{{urea}}}} = 1.754\% $
Molecular weight of cane sugar, ${{{M}}_{{{sug}}}} = 342{{g}}.{{mo}}{{{l}}^{ - 1}}$
The process of movement of solvent from low concentration to high concentration through a semipermeable membrane is termed as osmosis. This movement occurs when it attains an equilibrium. It can also be retarded by giving a pressure, which is known as osmotic pressure. This osmotic pressure depends only on the number of solute particles, thereby depends on the concentration.
As we have said, both cane sugar and urea have same concentration, i.e. ${{{C}}_{{{sug}}}} = {{{C}}_{{{urea}}}}$.
Concentration is expressed in ${{mol}}.{{{L}}^{ - 1}}$. Thus the above equation can be expressed as:
$\dfrac{{{{{n}}_{{{sug}}}}}}{{{{{V}}_{{{sug}}}}}} = \dfrac{{{{{n}}_{{{urea}}}}}}{{{{{V}}_{{{urea}}}}}}$, where ${{{n}}_{{{sug}}}}$ and ${{{V}}_{{{sug}}}}$ is the number of moles and volume of cane sugar respectively; and ${{{n}}_{{{urea}}}}$ and ${{{V}}_{{{urea}}}}$ is the number of moles and volume of urea respectively.
As we know, the number of moles can be obtained by dividing given mass, ${{w}}$ by molecular mass, ${{M}}$.
i.e. $\dfrac{{{{{w}}_{{{sug}}}}}}{{{{{M}}_{{{sug}}}}{{{V}}_{{{sug}}}}}} = \dfrac{{{{{w}}_{{{urea}}}}}}{{{{{M}}_{{{urea}}}}{{{V}}_{{{urea}}}}}}$
It is given that $10{{g}}$ of cane sugar is needed for $100{{mL}}$ of solution. So given mass of cane sugar, ${{{w}}_{{{sug}}}} = 10{{g}}$ and volume is ${{{V}}_{{{sug}}}} = 100{{mL = 0}}{{.1L}}$
Similarly ${{1}}{{.754g}}$ of urea is needed for $100{{mL}}$ of solution. So given mass of urea, ${{{w}}_{{{urea}}}} = 1.754{{g}}$ and volume is ${{{V}}_{{{urea}}}} = 100{{mL}} = 0.1{{L}}$.
Substituting all these values, we get
$\dfrac{{{{10g}}}}{{{{342g}}{{.mo}}{{{l}}^{ - 1}} \times {{0}}{{.1L}}}} = \dfrac{{{{1}}{{.754g}}}}{{{{{M}}_{{{urea}}}} \times {{0}}{{.1L}}}}$
On simplification, we get
${{{M}}_{{{urea}}}} = \dfrac{{{{1}}{{.754g}} \times {{342g}}{{.mo}}{{{l}}^{ - 1}} \times {{0}}{{.1L}}}}{{{{10g}} \times {{0}}{{.1L}}}} = 59.98{{g}}{{.mo}}{{{l}}^{ - 1}}$
Thus, the molecular mass of urea is $59.98{{g}}.{{mo}}{{{l}}^{ - 1}}$.

Note: We may confuse the terms hypertonic and hypotonic with isotonic. Three of them are different. Solution having less number of dissolved solutes than in other solutions is termed as a hypotonic solution. Solution having more number of dissolved solutes than in other solutions is termed as hypertonic solution.