A mixture of `4 g` of hydrogen and `8 g` of helium at (NTP) has a dencity about.

Options

A. `0.22 kg//m^3`
B. `0.62 kg//m^3`
C. `1.12 kg//m^3`
D. `0.13 kg//m^3`

Explanation

To find the density of a mixture of 4 g of hydrogen and 8 g of helium at Normal Temperature and Pressure (NTP), we can follow these steps:

### Step 1: Calculate the number of moles of hydrogen and helium.
- **For Hydrogen (H₂)**:
- Molar mass of hydrogen = 2 g/mol (since H₂ has a molar mass of 2 g/mol).
- Number of moles of hydrogen (N₁) = mass / molar mass = 4 g / 2 g/mol = 2 moles.

- **For Helium (He)**:
- Molar mass of helium = 4 g/mol.
- Number of moles of helium (N₂) = mass / molar mass = 8 g / 4 g/mol = 2 moles.

### Step 2: Calculate the total number of moles in the mixture.
- Total moles (N_total) = N₁ + N₂ = 2 moles (H₂) + 2 moles (He) = 4 moles.

### Step 3: Calculate the molar mass of the mixture.
- Molar mass (M) of the mixture can be calculated using the formula:
\[ M = \frac{N₁M₁ + N₂M₂}{N₁ + N₂} \]
where:
- M₁ = molar mass of hydrogen = 2 g/mol,
- M₂ = molar mass of helium = 4 g/mol.

- Substituting the values:
\[ M = \frac{(2 \text{ moles} \times 2 \text{ g/mol}) + (2 \text{ moles} \times 4 \text{ g/mol})}{2 + 2} = \frac{4 + 8}{4} = \frac{12}{4} = 3 \text{ g/mol}. \]

### Step 4: Calculate the density of the gas mixture.
- Using the ideal gas equation, we know:
\[ PV = nRT \]
Rearranging gives us:
\[ \text{Density} (\rho) = \frac{PM}{RT} \]
where:
- P = pressure at NTP = 1.013 × 10⁵ Pa,
- R = universal gas constant = 8.314 J/(mol·K),
- T = temperature at NTP = 273 K,
- M = molar mass of the mixture = 3 g/mol = 3 × 10⁻³ kg/mol (for SI units).

### Step 5: Substitute the values into the density formula.
- Substituting the values:
\[ \rho = \frac{(1.013 \times 10^5 \text{ Pa}) \times (3 \times 10^{-3} \text{ kg/mol})}{(8.314 \text{ J/(mol·K)}) \times (273 \text{ K})} \]

### Step 6: Calculate the density.
- First, calculate the denominator:
\[ 8.314 \times 273 \approx 2270.862 \text{ J/(mol·K)}. \]
- Now calculate the density:
\[ \rho = \frac{(1.013 \times 10^5) \times (3 \times 10^{-3})}{2270.862} \approx \frac{303.9}{2270.862} \approx 0.1337 \text{ kg/m}^3. \]

### Final Result:
- The density of the mixture at NTP is approximately **0.13 kg/m³**.