Question

A solution containing 10 g per \[{\text{d}}{{\text{m}}^{\text{3}}}\] of urea (m.w. = 60) is isotonic with a \[5\% \] solution of a non-volatile solute. The molecular mass of this non-volatile solute is:
A) 250 g \[{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\]
B) 300 g \[{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\]
C) 350 g \[{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\]
D) 200 g \[{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}\]

Answer 1

Hint: We know the concentration is the same for isotonic solutions. So, by equating the concentration of urea solution and the solution of a non-volatile solute we can get the molecular mass of the non-volatile solute.

Formula Used: \[{\text{concentration = }}\dfrac{{{\text{moles}}}}{{{\text{volume}}}}\]

Complete step by step answer:
We know that urea is non-volatile in nature:
Let us first calculate the concentration of the urea solution:
We know that the concentration is the number of moles divided by the volume of the solution
\[{\text{concentration = }}\dfrac{{{\text{moles}}}}{{{\text{volume}}}}\]
Now, we know that number of moles is given mass divided by the molecular mass of the substance. So, the formula can be modified as follows:
\[{\text{concentration = }}\dfrac{{{\text{given mass}}}}{{{\text{molecular mass X volume}}}}\]
Now, in the question, we are provided with the given mass of urea and molecular mass of urea.
So by substituting the value we get the concentration of urea as follows:
\[{{\text{C}}_{{\text{urea}}}} = \dfrac{{10{\text{g}}}}{{60{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ }} \times 1000{\text{c}}{{\text{m}}^{{\text{ - 3}}}}}}\] (equation 1)
Here, \[1{\text{ d}}{{\text{m}}^{\text{3}}} = 1000{\text{ c}}{{\text{m}}^{\text{3}}}\]
Now let us calculate the concentration of non-volatile solute:
For non-volatile solute, we are given with \[5\% \] a solution which means 5 grams of nonvolatile solute is dissolved in 100 \[{\text{c}}{{\text{m}}^{\text{3}}}\] solutions.
So the concentration of non-volatile solute becomes:
\[{{\text{C}}_{{\text{non - volatile}}}} = \dfrac{{5{\text{g}}}}{{{\text{M }} \times 100{\text{c}}{{\text{m}}^{{\text{ - 3}}}}}}\] (equation 2)
Here, M = molecular mass of the non-volatile solute.
Now, we know that both the solutions are isotonic which means:
\[{{\text{C}}_{{\text{urea}}}}{\text{ = }}{{\text{C}}_{{\text{non - volatile}}}}\]
By substituting the values form equation 1 and equation 2 we get:
\[\dfrac{{10{\text{g}}}}{{60{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ }} \times 1000{\text{c}}{{\text{m}}^{{\text{ - 3}}}}}} = \dfrac{{5{\text{g}}}}{{M{\text{ }} \times 100{\text{c}}{{\text{m}}^{{\text{ - 3}}}}}}\]
By taking all the values on one side we get:
\[{\text{M}} = \dfrac{{5{\text{g}} \times 60{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}{\text{ }} \times 1000{\text{c}}{{\text{m}}^{{\text{ - 3}}}}}}{{10{\text{g }} \times 100{\text{c}}{{\text{m}}^{{\text{ - 3}}}}}}\]
By solving the equation we get:
\[M = 300{\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}\]

Therefore, we can conclude that the correct answer to this question is option B.

Note: We must always focus on the conversion of the unit. Here \[{\text{d}}{{\text{m}}^{\text{3}}}\] has to be converted in \[{\text{c}}{{\text{m}}^{\text{3}}}\] because the \[5\% \] non-volatile solution means 5 grams of nonvolatile solute is dissolved in 100 \[{\text{c}}{{\text{m}}^{\text{3}}}\] of the solution and not 100 \[{\text{d}}{{\text{m}}^{\text{3}}}\] of the solution.