Question

A solution containing 3 g of solute A (M=60 g/mol) in 1L solution is isotonic with a solution containing 8.55 g

of solute B in 500 mL solution. What is the molar mass of B?​

Answer 1

Concept

• Osmotic Pressure, π = CRT.
• Two solutions are called isotonic if their osomotic pressure is same.

Solution

Since, Temperature is same. The solutions will be isotonic when :-

$C_A\:=\:C_B$

Let us find concentration of A.

$C_A\:=\: \dfrac{no.\:of\:moles\:of\:solute}{vol.\:of\:soln.\:in\:l} \\ \\ \\ \implies \:C_A\:=\: \dfrac{\frac{3}{60}}{1} \\ \\ \\ \implies C_A\:=\: \dfrac{1}{20}$

Let us assume molar mass of B be m.

Hence, concentration of B will be :-

$C_B\:=\: \dfrac{no.\:of\:moles\:of\:solute}{vol.\:of\:soln.\:in\:l} \\ \\ \\ \implies \:C_B\:=\: \dfrac{\frac{8.55}{m}}{0.5} \\ \\ \\ \implies C_B\:=\: \dfrac{17.1}{m}$

Equating concentration of both solutions :

$C_A\:=\:C_B \\ \\ \\ \implies \: \dfrac{1}{20}\:=\:\dfrac{17.1}{m} \\ \\ \\ \implies\:m \:=\: 342$

Hence, Molar mass of B is 342 g.

Answer 2

Osmotic Pressure, π = CRT.

Two solutions are called isotonic if their osomotic pressure is same.

Solution

Since, Temperature is same. The solutions will be isotonic when :-

Ca = Cb

Let us find concentration of A.

$\tt \begin{gathered} \tt \: C_A\:=\: \dfrac{no.\:of\:moles\:of\:solute}{vol.\:of\:soln.\:in\:l} \\ \\ \tt \implies \:C_A\:=\: \dfrac{\frac{3}{60}}{1} \\ \\ \tt \implies C_A\:=\: \dfrac{1}{20} \end{gathered} </p><p> \\ \\ \\ \tt C </p><p>A</p><p> </p><p> = </p><p> \tt vol.ofsoln \: .inl \: \: </p><p>no.of \: \: moles \: \: of \: \: solute</p><p> </p><p> </p><p> \\ \\ \\ \tt⟹C </p><p>A</p><p> </p><p> = </p><p>1</p><p>60</p><p>3</p><p> </p><p> </p><p> </p><p> </p><p>⟹C </p><p>A</p><p> </p><p> = </p><p>20</p><p>1</p><p>$

Let us assume molar mass of B be m.

Hence, concentration of B will be :-