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

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 Molar Mass of Each Gas
- **Hydrogen (H₂)** has a molar mass of approximately 2 g/mol.
- **Helium (He)** has a molar mass of approximately 4 g/mol.

### Step 2: Calculate the Number of Moles of Each Gas
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
- For hydrogen:
\[ n_1 = \frac{4 \text{ g}}{2 \text{ g/mol}} = 2 \text{ moles} \]
- For helium:
\[ n_2 = \frac{8 \text{ g}}{4 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Calculate the Average Molar Mass of the Mixture
The average molar mass (\(M_{mix}\)) of the mixture can be calculated using the formula:
\[ M_{mix} = \frac{m_1 \cdot M_1 + m_2 \cdot M_2}{m_1 + m_2} \]
Where:
- \(m_1\) = mass of hydrogen = 4 g
- \(M_1\) = molar mass of hydrogen = 2 g/mol
- \(m_2\) = mass of helium = 8 g
- \(M_2\) = molar mass of helium = 4 g/mol

Plugging in the values:
\[ M_{mix} = \frac{4 \cdot 2 + 8 \cdot 4}{4 + 8} = \frac{8 + 32}{12} = \frac{40}{12} \approx 3.33 \text{ g/mol} \]

### Step 4: Use the Ideal Gas Law to Calculate Density
The density (\( \rho \)) of an ideal gas can be calculated using the formula:
\[ \rho = \frac{PM_{mix}}{RT} \]
Where:
- \(P\) = pressure (at NTP, \(P = 1 \text{ atm} = 101.325 \text{ kPa}\))
- \(R\) = ideal gas constant = 8.314 J/(mol·K) = 0.0821 L·atm/(mol·K)
- \(T\) = temperature (at NTP, \(T = 273.15 \text{ K}\))

Using the appropriate units for density, we can convert pressure to the same unit as density:
\[ \rho = \frac{(101.325 \text{ kPa}) \cdot (3.33 \text{ g/mol})}{(8.314 \text{ J/(mol·K)}) \cdot (273.15 \text{ K})} \]
First, convert \(3.33 \text{ g/mol}\) to kg:
\[ 3.33 \text{ g/mol} = 0.00333 \text{ kg/mol} \]
Now substituting the values:
\[ \rho = \frac{(101.325 \times 10^3 \text{ Pa}) \cdot (0.00333 \text{ kg/mol})}{(8.314 \text{ J/(mol·K)}) \cdot (273.15 \text{ K})} \]

Calculating the denominator:
\[ 8.314 \cdot 273.15 \approx 2270.6 \text{ J/(mol)} \]

Now substituting back:
\[ \rho = \frac{(101325) \cdot (0.00333)}{2270.6} \approx \frac{337.2}{2270.6} \approx 0.1485 \text{ kg/m}^3 \]

### Final Answer
The density of the mixture is approximately \(0.1485 \text{ kg/m}^3\).