This law was proposed by Dalton in 1803 . According to this law , if two elements combune to form more than one compounds ,the masses of one element that combine with a fixed mass of the other element, are in the ratio of small whole number.
For exemple, hydrogen combines with oxygen to form two compounds ,namely, water and hydrogen peroxide.
Here, the masses of oxygen (i.e, 16g and 32g) which combine with a fixed mass of hydrogen (2g) bear a simple raio, i.e, 16 :32 or 1 :2
0.11 g of an oxide of nitrogen gives 56 Ml `N_(2)` at 273K and 1 atm. 0.15 g of another oxide of nitrogen gives 56mL `N_(2)` at 1 atm , 273K. Show that these data confirm the law of multiple proportion.

Answer

To solve the problem and confirm the law of multiple proportions using the provided data, we will follow these steps:

### Step 1: Understand the Given Data
We have two oxides of nitrogen that produce the same volume of nitrogen gas (N₂) under identical conditions (273 K and 1 atm). The masses of the oxides are:
- First oxide: 0.11 g
- Second oxide: 0.15 g

Both produce 56 mL of nitrogen gas.

### Step 2: Convert Volume of Nitrogen to Mass
Using the ideal gas law at standard temperature and pressure (STP), we know that 1 mole of gas occupies 22.4 L (or 22400 mL). The molar mass of nitrogen (N₂) is 28 g/mol.

To find the mass of nitrogen produced from 56 mL:
\[ \text{Mass of } N_2 = \left( \frac{56 \text{ mL}}{22400 \text{ mL}} \right) \times 28 \text{ g} = \frac{56 \times 28}{22400} = 0.07 \text{ g} \]

### Step 3: Determine the Mass of Oxygen in Each Oxide
For the first oxide:
- Mass of oxide = 0.11 g
- Mass of nitrogen = 0.07 g
- Therefore, mass of oxygen = mass of oxide - mass of nitrogen = 0.11 g - 0.07 g = 0.04 g

For the second oxide:
- Mass of oxide = 0.15 g
- Mass of nitrogen = 0.07 g (remains the same)
- Therefore, mass of oxygen = mass of oxide - mass of nitrogen = 0.15 g - 0.07 g = 0.08 g

### Step 4: Calculate the Ratio of Masses of Oxygen
Now we have the masses of oxygen that combine with a fixed mass of nitrogen (0.07 g):
- For the first oxide: mass of oxygen = 0.04 g
- For the second oxide: mass of oxygen = 0.08 g

The ratio of the masses of oxygen is:
\[ \text{Ratio} = \frac{0.04 \text{ g}}{0.08 \text{ g}} = \frac{1}{2} \]

### Step 5: Conclusion
The ratio of the masses of oxygen that combine with a fixed mass of nitrogen is 1:2, which is a simple whole number. This confirms Dalton's law of multiple proportions.

### Summary of the Steps:
1. Identify the given data (mass of oxides and volume of nitrogen).
2. Convert the volume of nitrogen to mass using the molar volume at STP.
3. Calculate the mass of oxygen in each oxide.
4. Determine the ratio of the masses of oxygen for the two oxides.
5. Conclude that the ratio is a simple whole number, confirming the law of multiple proportions.