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

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.