If a certain oxide of nitrogen weighing 0.11g gives 56 ml of nitrogen and another oxide of nitrogen weighing 0.15 g gives the same volume of nitrogen (both at STP), show that these results support the law of multiple proportions.

22400 ml of N2 = 28 gm so,

56 ml of N2 = 28 x 56/ 22400 = 0.07 gm

Here weight of oxide (A) =0.11 gm

After reduction the weight = 0.07gm

Loss of weight = 0.11-0.07 = 0.04gm (It is the amount of oxygen)--1.

Similarly weight of oxide(B) =0.15 gm

Weight after reduction = 0.07 gm

Hence, in case (A)

0.07 gm of nitrogen combines with 0.04 gm of oxygen

so, weight of nitrogen which would combine with 0.08 gm of oxygen should be equal to

= 0.07 x 0.08/ 0.04 = 0.14 gm--.---2

Here,in both the cases 1 and 2 the weight of nitrogen which combines with fixed amount of oxygen ( 0.04 gm) are 0.07 gm and 0.14 gm which are in the simple ratio of 2:1

Hence the oxides follow the law of multiple proportions.

Sir maine nitrogen ka mass fixed kr diya aur oxygen ka ratio nikala 1:2 aa rha hai kya mera answer shi hai kya please reply

Really liked the answer and I hope it will help other like me

Well, imagine these two oxides of nitrogen walking into a bar. The first one, weighing 0.11g, says to the bartender, "I'll have a glass of nitrogen, please." The bartender pours 56 ml of nitrogen and hands it over.

Then, in comes the second oxide of nitrogen, weighing 0.15g. It says to the bartender, "I'll have what he had, please." The bartender, being a good listener, pours the exact same 56 ml of nitrogen into a glass and gives it to the second oxide.

Now, let's analyze this situation. Both oxides of nitrogen requested the same amount of nitrogen, and both received the same volume. This suggests that there is a specific ratio between the mass of the oxides and the volume of nitrogen they produce.

In other words, the ratio between the masses of the two oxides (0.11g and 0.15g) is the same as the ratio between the volumes of nitrogen they produce (56 ml and 56 ml). This is precisely what the law of multiple proportions states!

So, based on these results, it seems like these two oxides of nitrogen are adhering to the law of multiple proportions, and they're keeping it consistent in the bar! Cheers to scientific laws and balanced chemical reactions!

To show that these results support the law of multiple proportions, we need to determine the ratio of the masses of the two oxides.

Let's start by setting up the equation for the first oxide of nitrogen:

Mass of first oxide (A): 0.11g
Volume of nitrogen produced from A: 56 mL

Now, for the second oxide of nitrogen:

Mass of second oxide (B): 0.15g
Volume of nitrogen produced from B: 56 mL

From the information given, we can assume that the volumes of nitrogen gas produced at STP are directly proportional to the moles of nitrogen gas produced. This is because at STP, 1 mole of any ideal gas occupies a volume of 22.4 L.

We can convert the volume of nitrogen gas (56 mL) to moles using the ideal gas law.

Using the ideal gas law:
PV = nRT

Since we are dealing with 1 mole of gas at STP (standard temperature and pressure - 273.15 K and 1 atm), we can simplify the equation:
V = nRT/P

Substituting values:
V = (1 mol x 0.0821 L·atm/(mol·K) x 273.15 K) / 1 atm
V = 22.4 L

Therefore, 1 mole of any ideal gas at STP occupies a volume of 22.4 L.

Now, let's calculate the moles of nitrogen produced from each oxide:

Moles of nitrogen from A: 56 mL / 22.4 L/mol = 2.5 x 10^-3 mol
Moles of nitrogen from B: 56 mL / 22.4 L/mol = 2.5 x 10^-3 mol

Now we can determine the molar ratios between both oxides.

Molar ratio = Moles of nitrogen produced / Molar mass of the oxide

For oxide A:
Molar ratio of A = (2.5 x 10^-3 mol) / (0.11 g)

For oxide B:
Molar ratio of B = (2.5 x 10^-3 mol) / (0.15 g)

Simplifying both ratios:
Molar ratio of A ≈ 0.02 mol/g
Molar ratio of B ≈ 0.02 mol/g

Since both molar ratios are approximately equal, we can conclude that the ratio of the masses of the two oxides is also approximately equal:

Mass ratio of A/B = 0.11 g / 0.15 g ≈ 0.73

This demonstrates the law of multiple proportions, as the ratio of the masses of the two substances (the oxides of nitrogen) can be expressed as small, whole numbers (in this case, approximately 3:4).

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