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Step-by-Step Solution
Step 1: Identify the Concept
We are dealing with the concept of osmotic pressure, where isotonic solutions have equal osmotic pressure. The formula for osmotic pressure is given by
$ \pi = C R T $,
where $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature.
Step 2: Express the Concentrations for Both Solutions
Since the solutions are isotonic (i.e., $\pi_1 = \pi_2$), their molar concentrations must be equal (assuming the same temperature and solvent):
$C_1 = C_2$.
(a) Concentration of the Unknown Substance
A 5.25% (w/w) solution means 5.25 g of the unknown substance in 100 g of the solution. Given the density is 1.0 g cm−3, 100 g of solution = 100 mL = 0.1 L of solution.
The number of moles of unknown substance in 100 mL is
$ \frac{5.25}{M} $,
where $M$ is the molar mass. Hence, the molar concentration (moles per liter) is
$ \frac{5.25 / M}{0.1} = \frac{52.5}{M} \text{ mol L}^{-1} $.
(b) Concentration of Urea Solution
A 1.5% (w/w) urea solution means 1.5 g of urea ($\text{molar mass} = 60 \text{ g mol}^{-1}$) in 100 g of solution. Again, 100 g of solution is 100 mL = 0.1 L. The number of moles of urea is
$ \frac{1.5}{60} = 0.025 \text{ moles} $.
Therefore, the molar concentration is
$ \frac{0.025}{0.1} = 0.25 \text{ mol L}^{-1} $.
Step 3: Equate the Concentrations
Since the two solutions are isotonic:
$$
C_1 = C_2 \implies \frac{52.5}{M} = 0.25
$$
Solving for $M$:
$$
M = \frac{52.5}{0.25} = 210 \text{ g mol}^{-1}.
$$
Step 4: State the Final Answer
The molar mass of the unknown substance is
$ 210 \text{ g mol}^{-1} $.
