To solve the problem, we need to find the molar mass of the non-volatile, non-electrolyte solute \(X\) that is iso-osmotic with a 10% sucrose solution. Here’s a step-by-step solution:
### Step 1: Determine the Molar Mass of Sucrose
The molecular formula of sucrose is \(C_{12}H_{22}O_{11}\).
To calculate its molar mass:
- Carbon (C): \(12 \times 12.01 \, \text{g/mol} = 144.12 \, \text{g/mol}\)
- Hydrogen (H): \(22 \times 1.008 \, \text{g/mol} = 22.176 \, \text{g/mol}\)
- Oxygen (O): \(11 \times 16.00 \, \text{g/mol} = 176.00 \, \text{g/mol}\)
Adding these together:
\[
\text{Molar mass of sucrose} = 144.12 + 22.176 + 176.00 = 342.30 \, \text{g/mol}
\]
### Step 2: Calculate the Number of Moles of Sucrose in 10% Solution
A 10% solution means there are 10 grams of sucrose in 100 mL of solution. For 1000 mL (1 L), there would be:
\[
\text{Mass of sucrose} = 10 \, \text{g} \times 10 = 100 \, \text{g}
\]
Now, calculate the number of moles of sucrose:
\[
\text{Number of moles of sucrose} = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{342.30 \, \text{g/mol}} \approx 0.292 \, \text{mol}
\]
### Step 3: Calculate the Osmotic Pressure of Sucrose Solution
Using the formula for osmotic pressure:
\[
\pi = CRT
\]
Where:
- \(C\) = concentration in mol/L (which is the number of moles in 1 L)
- \(R\) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \(T\) = temperature in Kelvin (assume standard temperature, 298 K)
Substituting the values:
\[
\pi = 0.292 \, \text{mol/L} \times 0.0821 \, \text{L·atm/(K·mol)} \times 298 \, \text{K} \approx 7.15 \, \text{atm}
\]
### Step 4: Determine the Molar Mass of Solute \(X\)
Now, we know that the 1.754% solution of \(X\) is iso-osmotic with the sucrose solution.
For a 1.754% solution, in 1000 mL (1 L), the mass of \(X\) is:
\[
\text{Mass of } X = 1.754 \, \text{g}
\]
To find the number of moles of \(X\):
\[
\text{Number of moles of } X = \frac{1.754 \, \text{g}}{M}
\]
Where \(M\) is the molar mass of \(X\).
### Step 5: Calculate the Osmotic Pressure of \(X\)
Using the same osmotic pressure formula:
\[
\pi = C \cdot R \cdot T
\]
Where \(C = \frac{1.754 \, \text{g}}{M} \text{ in 1 L}\):
\[
\pi = \frac{1.754}{M} \cdot 0.0821 \cdot 298
\]
### Step 6: Set the Osmotic Pressures Equal
Since the solutions are iso-osmotic:
\[
7.15 = \frac{1.754 \cdot 0.0821 \cdot 298}{M}
\]
### Step 7: Solve for \(M\)
Rearranging gives:
\[
M = \frac{1.754 \cdot 0.0821 \cdot 298}{7.15}
\]
Calculating this:
\[
M = \frac{43.298}{7.15} \approx 6.06 \, \text{g/mol}
\]
### Final Answer
The molar mass of solute \(X\) is approximately **6.06 g/mol**.
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