If 250 cm$$^3$$ of an aqueous solution containing 0.73g of a protein A is isotonic with one litre of another aqueous solution containing 1.65g of a protein B, at 298K, the ratio of the molecular masses of A and B is _________ $$\times 10^{-2}$$ (to the nearest integer).

For dilute solutions of non-electrolytes the van ’t Hoff relation for osmotic pressure is first stated:

$$\pi = \dfrac{n}{V}\,RT$$

where $$\pi$$ is the osmotic pressure, $$n$$ the number of moles of solute present in volume $$V$$, $$R$$ the gas constant and $$T$$ the absolute temperature.

Writing $$n = \dfrac{w}{M}$$ (mass over molar mass) we have

$$\pi = \dfrac{w}{M}\,\dfrac{RT}{V}$$

The two protein solutions are said to be isotonic, i.e. possess the same osmotic pressure at the common temperature 298 K. Hence

$$\pi_A = \pi_B$$

Substituting the expression for each solution:

$$\dfrac{w_A}{M_A}\,\dfrac{RT}{V_A} = \dfrac{w_B}{M_B}\,\dfrac{RT}{V_B}$$

The factors $$R$$ and $$T$$ are identical on both sides and therefore cancel out, leaving

$$\dfrac{w_A}{M_A V_A} = \dfrac{w_B}{M_B V_B}$$

Now we bring the molar masses together to form their ratio:

$$\dfrac{M_A}{M_B} = \dfrac{w_A V_B}{w_B V_A}$$

All the quantities on the right are known. We insert them one by one, taking care of the units. Volumes must be in the same unit, so we convert 250 cm3 to litres:

$$V_A = 250\;\text{cm}^3 = 0.250\;\text{L},\qquad V_B = 1\;\text{L}$$

The given masses are

$$w_A = 0.73\;\text{g},\qquad w_B = 1.65\;\text{g}$$

Substituting these four numbers:

$$\dfrac{M_A}{M_B} = \dfrac{0.73 \times 1.00}{1.65 \times 0.250}$$

Multiplying the denominator first:

$$1.65 \times 0.250 = 0.4125$$

and now carrying out the division:

$$\dfrac{0.73}{0.4125} = 1.770\;( \text{to three significant figures})$$

Thus

$$\dfrac{M_A}{M_B} \approx 1.77$$

To express this in the form requested in the question we rewrite 1.77 as a number multiplied by $$10^{-2}$$:

$$1.77 = 177 \times 10^{-2}$$

The blank to be filled is therefore 177 (to the nearest integer).

So, the answer is $$177$$.