The molality of the solution containing 70g of glucose dissolved in 2kg of water is ?

Options

A. 0.19m
B. 0.29m
C. 0.38m
D. 0.45m

Explanation

To calculate the molality of a solution containing 70g of glucose dissolved in 2kg of water, we can follow these steps:

### Step 1: Calculate the moles of glucose
First, we need to find the number of moles of glucose. The formula to calculate moles is:

\[ \text{Moles of solute} = \frac{\text{Mass of solute (g)}}{\text{Molar mass of solute (g/mol)}} \]

For glucose (C₆H₁₂O₆), the molar mass is approximately 180 g/mol. Given that we have 70g of glucose:

\[ \text{Moles of glucose} = \frac{70 \, \text{g}}{180 \, \text{g/mol}} \approx 0.3889 \, \text{moles} \]

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

### Step 3: Calculate the molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The formula for molality is:

\[ \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} \]

Substituting the values we have:

\[ \text{Molality} = \frac{0.3889 \, \text{moles}}{2 \, \text{kg}} \approx 0.1944 \, \text{mol/kg} \]

### Final Answer
Thus, the molality of the solution is approximately **0.19 mol/kg**.

---