An aqueous solution of sucrose (`C_(12)H_(22)O_(11)`) having a concentration of 34.2gram/ litre has an osmotic pressure of 2.38 atmospheres at `17^(@)`C. For an aqueous solution of glucose (`C_(6)H_(12)O_(6)`) to be isotonic with this solution , its concentration should be :

Options

A. (a) 34.2 gram per litre
B. (b) 17.1 gram per litre
C. (c) 18.0 gram per litre
D. (d) 36.0 gram per litre

Answer

To solve the problem step by step, we will use the relationship between osmotic pressure, concentration, and temperature.

### Step 1: Understand the Problem
We have an aqueous solution of sucrose with a known concentration and osmotic pressure. We need to find the concentration of an aqueous glucose solution that is isotonic with the sucrose solution.

### Step 2: Write the Formula for Osmotic Pressure
The osmotic pressure (\( \pi \)) of a solution can be expressed using the formula:
\[ \pi = i \cdot C \cdot R \cdot T \]
where:
- \( \pi \) = osmotic pressure
- \( i \) = Van't Hoff factor (which is 1 for non-electrolytes like glucose)
- \( C \) = concentration in mol/L
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

### Step 3: Convert Temperature to Kelvin
The temperature is given as \( 17^\circ C \). We need to convert this to Kelvin:
\[ T = 17 + 273 = 290 \, K \]

### Step 4: Set Up the Equation for Sucrose
For the sucrose solution, we know:
- \( \pi_1 = 2.38 \, \text{atm} \)
- Concentration of sucrose = 34.2 g/L

First, we need to calculate the molarity of the sucrose solution. The molar mass of sucrose (\( C_{12}H_{22}O_{11} \)) is:
\[ \text{Molar mass of sucrose} = (12 \times 12) + (22 \times 1) + (11 \times 16) = 342 \, g/mol \]
Now, we can find the number of moles of sucrose in 1 L:
\[ \text{Number of moles} = \frac{34.2 \, g}{342 \, g/mol} = 0.1 \, mol \]
Thus, the concentration \( C_1 \) of sucrose in mol/L is:
\[ C_1 = 0.1 \, mol/L \]

### Step 5: Calculate the Osmotic Pressure for Sucrose
Using the osmotic pressure formula:
\[ \pi_1 = i \cdot C_1 \cdot R \cdot T \]
Substituting the values:
\[ 2.38 = 1 \cdot 0.1 \cdot 0.0821 \cdot 290 \]
Calculating the right side:
\[ 0.1 \cdot 0.0821 \cdot 290 = 2.38 \, \text{atm} \]
This confirms that our calculations for sucrose are correct.

### Step 6: Set Up the Equation for Glucose
For the glucose solution, we want to find \( C_2 \) such that:
\[ \pi_2 = \pi_1 \]
Thus,
\[ 2.38 = 1 \cdot C_2 \cdot 0.0821 \cdot 290 \]
Solving for \( C_2 \):
\[ C_2 = \frac{2.38}{0.0821 \cdot 290} \]

### Step 7: Calculate \( C_2 \)
Calculating the denominator:
\[ 0.0821 \cdot 290 \approx 23.829 \]
Now substituting back:
\[ C_2 = \frac{2.38}{23.829} \approx 0.0999 \, mol/L \]

### Step 8: Convert Molarity to Grams per Liter
The molar mass of glucose (\( C_6H_{12}O_6 \)) is:
\[ \text{Molar mass of glucose} = (6 \times 12) + (12 \times 1) + (6 \times 16) = 180 \, g/mol \]
Now, we convert the molarity to grams per liter:
\[ \text{Concentration in g/L} = C_2 \times \text{Molar mass of glucose} = 0.0999 \, mol/L \times 180 \, g/mol \approx 18 \, g/L \]

### Final Answer
The concentration of the glucose solution that is isotonic with the sucrose solution is approximately **18 g/L**.

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