A `19.6 g` of a given gaseous sample contains `2.8 g` of molecules `(d = 0.75 g L^(-1))`, `11.2 g` of molecules `(d = 3 g L^(-1))` and `5.6 g` of molecules `(d = 1.5 g L^(-1))`. All density measurements are made at `STP`. Calculate the total number of molecules `(N)` present in the given sample. Report your answer in `10^(23) N`
Assume Avogardro's number as `6 xx 10^(23)`.

Explanation

To calculate the total number of molecules (N) present in the given gaseous sample, we will follow these steps:

### Step 1: Calculate the volume of each gas sample
Using the formula:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

1. For the first gas sample (2.8 g, density = 0.75 g/L):
\[ V_1 = \frac{2.8 \, \text{g}}{0.75 \, \text{g/L}} = 3.7333 \, \text{L} \]

2. For the second gas sample (11.2 g, density = 3 g/L):
\[ V_2 = \frac{11.2 \, \text{g}}{3 \, \text{g/L}} = 3.7333 \, \text{L} \]

3. For the third gas sample (5.6 g, density = 1.5 g/L):
\[ V_3 = \frac{5.6 \, \text{g}}{1.5 \, \text{g/L}} = 3.7333 \, \text{L} \]

### Step 2: Calculate the number of moles for each gas sample
Using the formula:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]
At STP, 1 mole of gas occupies 22.4 L. Therefore, we can calculate the moles for each sample as follows:

1. For the first gas sample:
\[ n_1 = \frac{2.8 \, \text{g}}{0.75 \, \text{g/L}} \times \frac{1}{22.4 \, \text{L}} = \frac{2.8}{0.75 \times 22.4} \, \text{moles} \]

2. For the second gas sample:
\[ n_2 = \frac{11.2 \, \text{g}}{3 \, \text{g/L}} \times \frac{1}{22.4 \, \text{L}} = \frac{11.2}{3 \times 22.4} \, \text{moles} \]

3. For the third gas sample:
\[ n_3 = \frac{5.6 \, \text{g}}{1.5 \, \text{g/L}} \times \frac{1}{22.4 \, \text{L}} = \frac{5.6}{1.5 \times 22.4} \, \text{moles} \]

### Step 3: Calculate the number of molecules for each gas sample
Using Avogadro's number \( N_A = 6 \times 10^{23} \):
\[ \text{Number of molecules} = \text{Moles} \times N_A \]

1. For the first gas sample:
\[ N_1 = n_1 \times N_A = \left(\frac{2.8}{0.75 \times 22.4}\right) \times 6 \times 10^{23} \]

2. For the second gas sample:
\[ N_2 = n_2 \times N_A = \left(\frac{11.2}{3 \times 22.4}\right) \times 6 \times 10^{23} \]

3. For the third gas sample:
\[ N_3 = n_3 \times N_A = \left(\frac{5.6}{1.5 \times 22.4}\right) \times 6 \times 10^{23} \]

### Step 4: Calculate the total number of molecules
\[ N = N_1 + N_2 + N_3 \]

### Final Calculation
After performing the calculations, we find:
- \( N_1 \approx 1 \times 10^{23} \)
- \( N_2 \approx 1 \times 10^{23} \)
- \( N_3 \approx 1 \times 10^{23} \)

Thus, the total number of molecules:
\[ N = 1 \times 10^{23} + 1 \times 10^{23} + 1 \times 10^{23} = 3 \times 10^{23} \]

### Answer
The total number of molecules present in the given sample is \( 3 \times 10^{23} \).

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