A 4 per cent solution of sucrose `(C_(12)H_(22)O_(11))` is isotonic with 3 percent solution of an unknown organic substance. Calculate the molecular weight of the unknown substance.

Explanation

To calculate the molecular weight of the unknown organic substance that is isotonic with a 4% sucrose solution, we can follow these steps:

### Step 1: Understand the definition of isotonic solutions
Isotonic solutions have the same osmotic pressure. Therefore, we can set the osmotic pressure of the sucrose solution equal to that of the unknown organic substance.

### Step 2: Calculate the osmotic pressure of the sucrose solution
The formula for osmotic pressure (π) is given by:
\[ \pi = \frac{W \cdot R \cdot T}{M \cdot V} \]
Where:
- \(W\) = mass of solute (in grams)
- \(R\) = universal gas constant (0.0821 L·atm/(K·mol))
- \(T\) = temperature in Kelvin (we will keep it as T for now)
- \(M\) = molecular weight of the solute (in g/mol)
- \(V\) = volume of the solution (in liters)

For the sucrose solution:
- \(W = 4 \, \text{g}\)
- \(M = 342 \, \text{g/mol}\) (molecular weight of sucrose)
- \(V = 0.1 \, \text{L}\)

Substituting these values into the osmotic pressure formula:
\[ \pi_1 = \frac{4 \cdot 0.0821 \cdot T}{342 \cdot 0.1} \]

### Step 3: Calculate the osmotic pressure of the unknown organic substance
For the unknown organic substance:
- \(W = 3 \, \text{g}\)
- \(M\) = unknown (what we need to find)
- \(V = 0.1 \, \text{L}\)

Substituting these values into the osmotic pressure formula:
\[ \pi_2 = \frac{3 \cdot 0.0821 \cdot T}{M \cdot 0.1} \]

### Step 4: Set the osmotic pressures equal to each other
Since the two solutions are isotonic, we have:
\[ \pi_1 = \pi_2 \]
Thus,
\[ \frac{4 \cdot 0.0821 \cdot T}{342 \cdot 0.1} = \frac{3 \cdot 0.0821 \cdot T}{M \cdot 0.1} \]

### Step 5: Simplify the equation
We can cancel \(0.0821\) and \(T\) from both sides:
\[ \frac{4}{342 \cdot 0.1} = \frac{3}{M \cdot 0.1} \]

### Step 6: Cross-multiply to solve for M
Cross-multiplying gives us:
\[ 4 \cdot M \cdot 0.1 = 3 \cdot 342 \cdot 0.1 \]
Dividing both sides by \(0.1\):
\[ 4M = 3 \cdot 342 \]

### Step 7: Calculate M
Now, solving for \(M\):
\[ M = \frac{3 \cdot 342}{4} \]
Calculating the right side:
\[ M = \frac{1026}{4} = 256.5 \, \text{g/mol} \]

### Conclusion
The molecular weight of the unknown organic substance is **256.5 g/mol**.

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