1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of $$2.42 \times 10^{-3}$$ bar.
The molar mass of the biopolymer is _________ $$\times 10^4$$ g mol$$^{-1}$$. (Round off to the Nearest Integer)
[Use: R = 0.083 L bar mol$$^{-1}$$ K$$^{-1}$$]

For dilute solutions, osmotic pressure is given by the van’t Hoff equation:

$$\pi = C\,R\,T$$

where $$\pi$$ is the osmotic pressure, $$C$$ is the molar concentration $$\left(\dfrac{n}{V}\right)$$, $$R$$ is the gas constant and $$T$$ is the absolute temperature.

We write the number of moles as $$n = \dfrac{m}{M}$$, with $$m$$ the mass of solute and $$M$$ its molar mass. Substituting $$n$$ in the above relation, we get

$$\pi = \dfrac{m}{M\,V}\;R\,T$$

Re-arranging to isolate $$M$$,

$$M = \dfrac{m\,R\,T}{\pi\,V}$$

Now we substitute every given quantity, taking care of units:

Mass of biopolymer, $$m = 1.46\ \text{g}$$

Gas constant, $$R = 0.083\ \text{L bar mol}^{-1}\text{K}^{-1}$$

Temperature, $$T = 300\ \text{K}$$

Osmotic pressure, $$\pi = 2.42 \times 10^{-3}\ \text{bar}$$

Volume of solution, $$V = 100\ \text{mL} = 0.100\ \text{L}$$

Substituting,

$$M = \dfrac{1.46 \times 0.083 \times 300}{\left(2.42 \times 10^{-3}\right)\times 0.100}$$

First we evaluate the numerator:

$$0.083 \times 300 = 24.9$$

$$1.46 \times 24.9 = 36.354$$

Hence the numerator is $$36.354$$.

Now we evaluate the denominator:

$$2.42 \times 10^{-3} \times 0.100 = 2.42 \times 10^{-4}$$

So,

$$M = \dfrac{36.354}{2.42 \times 10^{-4}}$$

Dividing,

$$M = 1.50 \times 10^{5}\ \text{g mol}^{-1}$$

Expressing this in the requested form $$\left(\_\_\_\_\_ \times 10^{4}\ \text{g mol}^{-1}\right)$$, we write

$$1.50 \times 10^{5}\ \text{g mol}^{-1} = 15 \times 10^{4}\ \text{g mol}^{-1}$$

Rounding to the nearest integer, the blank is filled by $$15$$.

So, the answer is $$15$$.