Question

At $27^\circ C$, a $1.2\% $ solution ($wt/vol.$) of glucose is isotonic with $4.0 g/L$ of solution of solute $X$. Find the molar mass of $X$, if the molar mass of glucose is $180$. ($R = 0.082 Latm/molK,$ Molar mass of glucose $ = 180 g/mol$).

Answer 1

Hint: Two solutions are said to be isotonic solutions if these solutions have same osmotic pressure across a semipermeable membrane or when two solutions have same concentration of solutes across a semipermeable membrane.

Complete step by step answer:
As we know that the isotonic solutions are those solutions which have the same osmotic pressure across a semipermeable membrane. The osmotic pressure is defined as the minimum pressure that is required to be applied to a solution to prevent the flow of its pure solvent across a semipermeable membrane. Semipermeable membrane only allows solvent molecules to pass through it. The osmotic pressure is represented as ($\pi $). The formula of osmotic pressure can be:
$\pi = iCRT = \dfrac{{inRT}}{V} = \dfrac{{imRT}}{{MV}}$ $ - (1)$
Where, $\pi = $ osmotic pressure
$i = $ van’t Hoff factor
$C = $ molar concentration of solute in a solution. It is given as:
 ($C = \dfrac{{{\text{no}}{\text{. of moles of solute}}}}{{{\text{Volume of solution}}}} = \dfrac{n}{V}$ and also, number of moles is $n = \dfrac{m}{M} = \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$)
$R = $ Gas constant
$T = $ Temperature.
As given in question that glucose and $X$ solution are isotonic so they will have same osmotic pressure that is
${\pi _{glucose}} = {\pi _X}$
Now for glucose, the van’t Hoff factor($i$) is one. This is because the Van’t Hoff factor is the ratio of dissociated molecules to the associated molecules but glucose is a non electrolyte and it does not dissociate.
Glucose has $1.2\% $ solution ($wt/vol.$) which means $1.2g$of glucose is in $100L$. Therefore, it will have mass of $0.012g$ in $1L$ of solution. Temperature in kelvin will be $27 + 273 = 300K$.
Now, from equation $ - (1)$
$
  {\pi _{glucose}} = \dfrac{{imRT}}{{MV}} \\
   = \dfrac{{1 \times 0.012 \times 0.082 \times 300}}{{180 \times 1}} \\
   = 0.00164atm \\
 $
For solution $X$, we have $4g$ of mass dissolved in $1L$ of volume and temperature is same as $300K$.
Now,
$
  {\pi _X} = \dfrac{{imRT}}{{MV}} \\
   = \dfrac{{1 \times 4 \times 0.082 \times 300}}{{M \times 1}} \\
   = \dfrac{{98.4}}{M} \\
 $
As we know that,
${\pi _{glucose}} = {\pi _X}$
Therefore, $
  0.00164 = \dfrac{{98.4}}{M} \\
  M = \dfrac{{98.4}}{{0.00164}} \\
  M = 60000g \\
 $

Hence, the required molar mass of $X$ is $60000g$.

Note: Remember that a semipermeable membrane is a type of membrane which allows only certain types of molecules to pass through it or in general it allows only the passage of solvent molecules but it does not allow solute molecules to pass through it.