The temperature at which 10% aqueous solution (w/v) of glucose will exhibit the osmotic pressure of 14 bar is -

Options

A. `307.3^@C`
B. `307.3 K`
C. `34K`
D. `-34^@C`

Explanation

To find the temperature at which a 10% (w/v) aqueous solution of glucose exhibits an osmotic pressure of 14 bar, we can use the formula for osmotic pressure:

\[ \pi = C \cdot R \cdot T \]

Where:
- \(\pi\) = osmotic pressure (in bar)
- \(C\) = concentration (in mol/L)
- \(R\) = universal gas constant (0.0821 L·bar/K·mol)
- \(T\) = temperature (in Kelvin)

### Step 1: Determine the concentration \(C\)

A 10% (w/v) solution means there are 10 grams of glucose in 100 mL of solution. To find the concentration in moles per liter (mol/L), we first need to calculate the number of moles of glucose.

The molecular mass of glucose (C₆H₁₂O₆) is approximately 180 g/mol.

\[ \text{Number of moles of glucose} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{10 \text{ g}}{180 \text{ g/mol}} = \frac{1}{18} \text{ mol} \]

Now, since the volume of the solution is 100 mL, we convert it to liters:

\[ \text{Volume in L} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]

Now we can calculate the concentration \(C\):

\[ C = \frac{\text{Number of moles}}{\text{Volume (L)}} = \frac{\frac{1}{18} \text{ mol}}{0.1 \text{ L}} = \frac{1}{1.8} \text{ mol/L} \approx 0.5556 \text{ mol/L} \]

### Step 2: Substitute values into the osmotic pressure formula

Now we can substitute the values into the osmotic pressure formula:

\[ 14 \text{ bar} = 0.5556 \text{ mol/L} \cdot 0.0821 \text{ L·bar/K·mol} \cdot T \]

### Step 3: Solve for \(T\)

Rearranging the equation to solve for \(T\):

\[ T = \frac{14 \text{ bar}}{0.5556 \text{ mol/L} \cdot 0.0821 \text{ L·bar/K·mol}} \]

Calculating the denominator:

\[ 0.5556 \cdot 0.0821 \approx 0.0456 \]

Now substituting this back into the equation for \(T\):

\[ T = \frac{14}{0.0456} \approx 307.3 \text{ K} \]

### Conclusion

The temperature at which a 10% aqueous solution of glucose will exhibit an osmotic pressure of 14 bar is approximately **307.3 K**.

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