A 5% solution of glucose (molar mass 180) is isotonic with a 2.5% solution of substance X at the same temperature. The molar mass of X is

Options

A. 90
B. 45
C. 180
D. 360

Explanation

To find the molar mass of substance X that is isotonic with a 5% glucose solution, we can follow these steps:

### Step 1: Understand the 5% Glucose Solution
A 5% solution of glucose means that there are 5 grams of glucose in 100 cc (or mL) of solution.

### Step 2: Calculate the Molarity of the Glucose Solution
To find the molarity (C1) of the glucose solution, we use the formula:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
First, we need to calculate the number of moles of glucose:
\[ \text{Moles of glucose} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \text{ g}}{180 \text{ g/mol}} = \frac{5}{180} \text{ mol} \]
Since the volume of the solution is 100 cc (which is 0.1 L), we can find the molarity:
\[ C_1 = \frac{\frac{5}{180}}{0.1} = \frac{5}{18} \text{ mol/L} \]

### Step 3: Understand the Isotonic Condition
Since the solutions are isotonic, the molarity of the glucose solution (C1) is equal to the molarity of the substance X (C2):
\[ C_1 = C_2 \]

### Step 4: Calculate the Molarity of Substance X
For the 2.5% solution of substance X, we have:
\[ \text{Mass of X} = 2.5 \text{ g in 100 cc (or 0.1 L)} \]
Let the molar mass of substance X be M. The number of moles of substance X is:
\[ \text{Moles of X} = \frac{2.5 \text{ g}}{M} \]
Thus, the molarity of substance X (C2) is:
\[ C_2 = \frac{\frac{2.5}{M}}{0.1} = \frac{2.5}{0.1M} = \frac{25}{M} \text{ mol/L} \]

### Step 5: Set the Molarities Equal
Since C1 = C2, we set the two expressions equal to each other:
\[ \frac{5}{18} = \frac{25}{M} \]

### Step 6: Solve for M
Cross-multiply to solve for M:
\[ 5M = 18 \times 25 \]
\[ 5M = 450 \]
\[ M = \frac{450}{5} = 90 \text{ g/mol} \]

### Conclusion
The molar mass of substance X is 90 g/mol.