Which of the following has lowest weight ?

Options

A. `6.023 xx 10^(22)` molecules of glucose
B. 18 ml of water at `4^(@)C`
C. 11200 ml of `CH_(4)` at STP
D. 5.6 litre of `CO_(2)` at STP

Answer

To determine which of the given options has the lowest weight, we will calculate the mass for each option step by step.

### Step 1: Calculate the mass of 6.023 x 10^22 molecules of glucose
1. **Calculate the number of moles**:
\[ \text{Moles} = \frac{\text{Number of molecules}}{\text{Avogadro's number}} = \frac{6.023 \times 10^{22}}{6.023 \times 10^{23}} = 0.1 \text{ moles} \]
2. **Calculate the mass**:
- Molar mass of glucose (C6H12O6) = 180 g/mol
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} = 0.1 \times 180 = 18 \text{ grams} \]

### Step 2: Calculate the mass of 18 ml of water at 4 °C
1. **Density of water at 4 °C** = 1 g/ml
2. **Calculate the mass**:
\[ \text{Mass} = \text{Volume} \times \text{Density} = 18 \text{ ml} \times 1 \text{ g/ml} = 18 \text{ grams} \]

### Step 3: Calculate the mass of 11,200 ml of CH4 at 80 °C
1. **Calculate the number of moles**:
\[ \text{Moles} = \frac{\text{Volume}}{22.4 \text{ L}} = \frac{11.2 \text{ L}}{22.4 \text{ L}} = 0.5 \text{ moles} \]
2. **Calculate the mass**:
- Molar mass of methane (CH4) = 16 g/mol
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} = 0.5 \times 16 = 8 \text{ grams} \]

### Step 4: Calculate the mass of 5.6 liters of CO2 at STP
1. **Calculate the number of moles**:
\[ \text{Moles} = \frac{\text{Volume}}{22.4 \text{ L}} = \frac{5.6 \text{ L}}{22.4 \text{ L}} = 0.25 \text{ moles} \]
2. **Calculate the mass**:
- Molar mass of carbon dioxide (CO2) = 44 g/mol
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} = 0.25 \times 44 = 11 \text{ grams} \]

### Summary of Masses
1. 6.023 x 10^22 molecules of glucose: 18 grams
2. 18 ml of water: 18 grams
3. 11,200 ml of CH4: 8 grams
4. 5.6 liters of CO2: 11 grams

### Conclusion
The option with the lowest weight is **11,200 ml of CH4**, which weighs **8 grams**.

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