The molality of the solution containing 75g of glucose dissolved in 3kg of water is ?

Options

A. 0.138m
B. 0.256m
C. 0.013m
D. 0.21m

Explanation

To find the molality of the solution containing 75g of glucose dissolved in 3kg of water, we can follow these steps:

### Step 1: Calculate the moles of glucose
To find the moles of glucose, we use the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]
The molar mass of glucose (C6H12O6) is approximately 180 g/mol.

Given:
- Mass of glucose = 75 g
- Molar mass of glucose = 180 g/mol

Calculating the moles of glucose:
\[ \text{Moles of glucose} = \frac{75 \, \text{g}}{180 \, \text{g/mol}} = 0.4167 \, \text{moles} \]

### Step 2: Calculate the mass of the solvent in kg
The mass of the solvent (water) is given as 3 kg.

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ \text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
Substituting the values we have:
\[ \text{Molality} = \frac{0.4167 \, \text{moles}}{3 \, \text{kg}} = 0.1389 \, \text{mol/kg} \]

### Final Answer
The molality of the solution is approximately **0.139 mol/kg**.

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