Question

A $1\% $aqueous solution $(\omega /\upsilon )$of a certain substance is isotonic with a $3\% $solution of dextrose i.e., glucose (molar mass $180$) at a temperature. The molar mass of the substance is:
(A) 60
(B) 120
(C) 180
(D) 360

Answer 1

Hint: When osmotic pressure of two solutions are the same, then they are said to be isotonic. Osmotic pressure is a colligative property which depends upon the number of solute particles in solution.
We can use the Formula:
For isotonic solution
${\pi _1} = {\pi _2}$where $\pi $osmotic pressure and $\pi = CRT.$

Step by step answer:Osmotic pressure is defined as minimum pressure that must be applied to a solution to stop the flow of solvent molecules through a semipermeable membrane.
Osmotic pressure represent by $\pi $its formula is $\pi = CRT$
Where, $C = $concentration of solution
$R = $gas constant
$T = $absolute pressure
$C = $concentration in mole/l
$C = \dfrac{n}{\upsilon }$or
$C = \dfrac{\omega }{{m \times \upsilon }}$
$\omega = $mass of substance
$m = $molecular mass
$\upsilon = $volume of solution
Let osmotic pressure of solution of unknown substance
${\pi _1} = {C_1}RT$or ${\pi _1} = \dfrac{{{\omega _1}}}{{{m_1} \times {\upsilon _1}}}$
${\omega _1} = $weight of substance
${m_1} = $molecular weight
${\upsilon _1} = $volume of solution
Osmotic pressure of solution of glucose
${\pi _2} = {C_2}RT$or ${\pi _2} = \dfrac{{{\omega _2}}}{{{m_2} \times {\upsilon _2}}}$
${\omega _2} = $weight of glucose
${m_2} = $molecular weight of glucose
${\upsilon _2} = $volume of solution
Since $1\% $of solution unknown substance is given
$\therefore {\omega _1} = 1gm$ ${\upsilon _1} = 100$ ${m_1} = ?$
$3\% $of glucose solution is given
$\therefore {\omega _2} = 3gm$ ${\upsilon _2} = 100$ ${m_2} = 180$
For Isotonic solution
${\pi _1} = {\pi _2}$
$\dfrac{{{\omega _1}}}{{{m_1} \times {\upsilon _1}}} = \dfrac{{{\omega _2}}}{{{m_2} \times {\upsilon _1}}}$__________ (1)
Substituting above value in equation (1) we get,
$\dfrac{1}{{{m_1} \times 100}} = \dfrac{3}{{180 \times 100}}$
$\Rightarrow 3 \times {m_1} = \dfrac{{180 \times 100}}{{100}}$
$\Rightarrow {m_1} = \dfrac{{180}}{3}$
$\Rightarrow {m_1} = 60$
$\therefore $molecular weight of unknown substance$ = 60.$
Therefore, from the above explanation the correct option is (A) $60.$

Note: Osmotic pressure of a salt solution in water exit pressure against a semi permeable membrane.
It depends on the number of solute particles in solution.
If they are different, they show different osmotic pressure.