$\text{At}\mathrm{ 300}K,\text{an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the}$

$\text{height}\left(h\right)\text{of the solution}(\text{density}=1.00g{\mathrm{cm}}^{-3})\text{where}\mathrm{ h}\text{ is equal to}\mathrm{ 2.00}\mathrm{cm}.\text{If the concentration of the}$

$\text{dilute solution of the macromolecule is}\mathrm{ 2\; .00}g{\mathrm{dm}}^{-3},\text{the molar mass of the macromolecule is calculated}$

$\text{to b e}X\times {10}^{4}g{\mathrm{mol}}^{-1}.\text{The value of}\mathrm{ X }\text{is}\text{\_\_\_\_\_}$
$\text{Use: Universal gas constant}\left(R\right)=8.3J{K}^{-1}{\mathrm{mol}}^{-1}\text{and acceleration due to gravity}\left(g\right)=10\frac{m}{s^2}$

The correct answer is 2.49.

To find the value of $X$, we can relate the osmotic pressure of the solution to its height and concentration using the laws of osmotic pressure.

Step 1: Calculate the osmotic pressure ($\Pi$) of the solution
The osmotic pressure exerted by a column of liquid is given by the hydrostatic pressure formula:
$\Pi =\rho gh$
Where:
• $\rho$ is the density of the solution = $1.00{\text{g cm}}^{-3}=1000{\text{kg m}}^{-3}$
• $g$ is the acceleration due to gravity = $10{\text{m s}}^{-2}$
• $h$ is the height of the column = $2.00\text{cm}=0.02\text{m}$

Substituting these values into the hydrostatic equation (in SI units):
$\Pi =1000\text{kg}{\text{m}}^{-3}\times 10\text{m}{\text{s}}^{-2}\times 0.02\text{m}$
$\Pi =200{\text{N m}}^{-2}\text{(or Pa)}$

Step 2: Relate osmotic pressure to molar concentration
According to the van 't Hoff equation for dilute solutions:
$\Pi =CRT$
Where:
• $C$ is the molar concentration in ${\text{mol m}}^{-3}$
• $R$ is the gas constant = $8.3{\text{J K}}^{-1}{\text{mol}}^{-1}$
• $T$ is the temperature = $300\text{K}$

The molar concentration $C$ can be expressed as:
$C=\frac{c}{M}$
Where $c$ is the concentration in mass per unit volume (${\text{g m}}^{-3}$) and $M$ is the molar mass (${\text{g mol}}^{-1}$).
Given the concentration is $2.00{\text{g dm}}^{-3}$:
Since $1{\text{dm}}^3={10}^{-3}{\text{m}}^3$, we have:
$c=\frac{2.00\text{g}}{{10}^{-3}{\text{m}}^3}=2000{\text{g m}}^{-3}$

Thus, substituting $C=\frac{2000}{M}$ into the van 't Hoff equation:
$\Pi =\frac{2000}{M}\times R\times T$

Step 3: Solve for the molar mass ($M$)
Substitute the values of $\Pi$, $R$, and $T$:
$200=\frac{2000}{M}\times 8.3\times 300$
Rearranging the equation to solve for $M$:
$M=\frac{2000\times 8.3\times 300}{200}$
$M=10\times 8.3\times 300$
$M=24900{\text{g mol}}^{-1}$

Expressing this in scientific notation:
$M=2.49\times {10}^4{\text{g mol}}^{-1}$

Comparing this to the given expression $X\times {10}^4{\text{g mol}}^{-1}$, we find:
$X=2.49$