Question

: A solution containing \[8.6{\text{ g }}{{\text{L}}^{-1} }\] of urea is isotonic with a \[5\% {\text{ (w/v)}}\] solution of unknown solute. The molar mass of solute will be
A.\[{\text{348}}{\text{.9 g mo}}{{\text{l}}^{-1} }\]
B.\[{\text{174}}{\text{.5 g mo}}{{\text{l}}^{-1} }\]
C.\[{\text{87}}{\text{.3 g mo}}{{\text{l}}^{-1} }\]
D.\[{\text{34}}{\text{.89 g mo}}{{\text{l}}^{-1} }\]

Answer 1

Hint:Isotonic solutions are those solutions which have the same concentration or we can say they have the same molarity. We will calculate the molarity of both the solution and equate them. We will get the molar mass of the unknown solute.
Formula used:
\[{\text{Molarity }} = \dfrac{{\text{n}}}{{{\text{V in mL}}}} \times 1000\] here n is number of moles and V is volume of solution in millilitre.
\[{\text{Number of moles }} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}\]

Complete step by step answer:
Isotonic solutions are the solutions which have the same concentration.
We have been given the question that concentration of one solution is \[8.6{\text{ g }}{{\text{L}}^{-1} }\]. But molarity is calculated in mol per litre and not in gram per litre. We will convert it into moles by dividing it with the molar mass of urea that is 60. Hence the molarity becomes, \[\dfrac{{8.6}}{{60}}{\text{ mol }}{{\text{L}}^{-1} }\]
Now we will calculate the concentration on the other solution. \[5\% {\text{ (w/v)}}\] Mean that 5 gram of unknown solute is present in 100 mL of solution. We will calculate the molarity using the formula of molarity as:
\[{\text{Molarity }} = \dfrac{5}{{{\text{molar mass}} \times {\text{100}}}} \times 1000\]
\[ \Rightarrow {\text{Molarity }} = \dfrac{{50}}{{{\text{molar mass}}}}\]
Now since both the solutions are isotonic hence we will equate the concentration of both and hence we will get:
$\Rightarrow$\[\dfrac{{8.6}}{{60}}{\text{ mol }}{{\text{L}}^{-1} } = \dfrac{{50}}{{{\text{molar mass}}}}\]
Rearranging the above equation we will get molar mass as:
\[{\text{molar mass }} = 348.9{\text{ g}}/{\text{L}}\]

Hence, the correct option is A.

Note:
Solutions on the basis of concentration can be classified as hypertonic, hypotonic and isotonic solutions. The solution having more number of moles of solute present per litre of solution with respect to the reference solution are called hypertonic solution and the solution which have less concentration than other are called hypotonic solution. Molarity is the concentration term and w/v is percentage composition specifically called as percentage by volume.