At `27^(@)C,a1.2%` solution (`wt.//"vol.")` of glucose is isotonic with `4.0g//"litre"` of urea solution. Find the molar mass of urea, if the molar mass of glucose is `180`. Also report the osmotic pressure of solution if `100mL` of each are mixed at `27^(@)C`. (`R=0.082 L` atm `"mol"^(-1)K^(-1)`, Molar mass of glucose `=180g//"mole"`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Concept of Isotonic Solutions
Isotonic solutions have the same osmotic pressure. Therefore, we can set the osmotic pressures of the glucose solution and the urea solution equal to each other.

### Step 2: Calculate the Concentration of Glucose
Given that we have a 1.2% (w/v) solution of glucose, this means there are 1.2 grams of glucose in 100 mL of solution.

To find the concentration in moles per liter (M), we use the formula:
\[ \text{Concentration} (C_1) = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)} \times \text{volume of solution (L)}} \]
Substituting the values for glucose:
\[ C_1 = \frac{1.2 \, \text{g}}{180 \, \text{g/mol} \times 0.1 \, \text{L}} = \frac{1.2}{18} = 0.06667 \, \text{mol/L} \]

### Step 3: Calculate the Concentration of Urea
We know that the urea solution has 4 g of urea per liter. The concentration of urea (C2) can be expressed as:
\[ C_2 = \frac{4 \, \text{g}}{M_2 \, \text{g/mol} \times 1 \, \text{L}} = \frac{4}{M_2} \]
Where \(M_2\) is the molar mass of urea that we need to find.

### Step 4: Set the Concentrations Equal
Since the solutions are isotonic, we can set \(C_1 = C_2\):
\[ 0.06667 = \frac{4}{M_2} \]
Now, we can solve for \(M_2\):
\[ M_2 = \frac{4}{0.06667} \approx 60 \, \text{g/mol} \]

### Step 5: Calculate the Total Osmotic Pressure
Next, we need to calculate the total osmotic pressure when 100 mL of each solution is mixed. First, we find the number of moles of each solute.

**For Glucose:**
\[ \text{Moles of glucose} = C_1 \times V = 0.06667 \, \text{mol/L} \times 0.1 \, \text{L} = 0.006667 \, \text{mol} \]

**For Urea:**
Using the molar mass we calculated:
\[ \text{Moles of urea} = \frac{4 \, \text{g}}{60 \, \text{g/mol}} = 0.06667 \, \text{mol} \]

### Step 6: Calculate Total Moles and Total Volume
Total moles when both solutions are mixed:
\[ \text{Total moles} = 0.006667 + 0.06667 = 0.073337 \, \text{mol} \]
Total volume after mixing:
\[ V_{\text{total}} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L} \]

### Step 7: Calculate the Osmotic Pressure
Using the formula for osmotic pressure:
\[ \pi = \frac{nRT}{V} \]
Where:
- \(n = 0.073337 \, \text{mol}\)
- \(R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1}\)
- \(T = 27^\circ C = 300 \, \text{K}\)

Substituting the values:
\[ \pi = \frac{0.073337 \times 0.082 \times 300}{0.2} = \frac{1.798}{0.2} = 8.99 \, \text{atm} \]

### Final Answers
- Molar mass of urea: **60 g/mol**
- Osmotic pressure of the mixed solution: **8.99 atm**