A solution containing 10.2 g of glycrine per litre is found to be isotonic with 2% solution of glucose `("molar mass"=180 g mol^(-1))`. Calculate the molar mass of glycrine.

Options

A. 91.8 g
B. 1198 g
C. 83.9 g
D. 890.3 g

Explanation

To solve the problem of finding the molar mass of glycerin given that it is isotonic with a 2% glucose solution, we can follow these steps:

### Step 1: Understand the concept of isotonic solutions
Isotonic solutions have the same osmotic pressure. Therefore, the osmotic pressure of the glycerin solution will be equal to that of the glucose solution.

### Step 2: Write the formula for osmotic pressure
The osmotic pressure (π) can be expressed as:
\[ \pi = C \cdot R \cdot T \]
where \(C\) is the molar concentration, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin. Since both solutions are isotonic, we can set the osmotic pressures equal to each other:
\[ C_{\text{glycerin}} = C_{\text{glucose}} \]

### Step 3: Calculate the concentration of glucose
Given that we have a 2% solution of glucose, we can calculate the concentration in moles per liter.

1. A 2% solution means there are 2 grams of glucose in 100 mL of solution.
2. To find the amount in 1 L (1000 mL), we have:
\[ \text{Mass of glucose in 1 L} = 2 \, \text{g} \times 10 = 20 \, \text{g} \]
3. The molar mass of glucose is 180 g/mol. Therefore, the number of moles of glucose in 1 L is:
\[ \text{Moles of glucose} = \frac{20 \, \text{g}}{180 \, \text{g/mol}} = \frac{1}{9} \, \text{mol} \]
4. The concentration of glucose (C) is:
\[ C_{\text{glucose}} = \frac{1}{9} \, \text{mol/L} \]

### Step 4: Calculate the concentration of glycerin
Since the osmotic pressures are equal, we can equate the concentrations:
\[ C_{\text{glycerin}} = C_{\text{glucose}} = \frac{1}{9} \, \text{mol/L} \]

### Step 5: Relate the concentration of glycerin to its molar mass
The concentration of glycerin can also be expressed in terms of its mass and molar mass (M):
\[ C_{\text{glycerin}} = \frac{\text{mass of glycerin}}{\text{molar mass of glycerin} \times \text{volume in L}} \]
Given that the mass of glycerin is 10.2 g and the volume is 1 L:
\[ \frac{10.2 \, \text{g}}{M \times 1 \, \text{L}} = \frac{1}{9} \, \text{mol/L} \]

### Step 6: Solve for the molar mass of glycerin
Rearranging the equation gives:
\[ 10.2 = \frac{M}{9} \]
Multiplying both sides by 9:
\[ M = 10.2 \times 9 = 91.8 \, \text{g/mol} \]

### Final Answer
The molar mass of glycerin is \(91.8 \, \text{g/mol}\).

---