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Question

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All gas volumes were measured at the same temperature and pressure.

What could be the identity of J?

1) ${{C}_{2}}{{H}_{6}}{{N}_{2}}$

2) ${{C}_{2}}{{H}_{3}}N$

3) ${{C}_{2}}{{H}_{7}}N$

(a)- 1, 2, and 3 are correct

(b)- 1 and 2 only are correct

(c)- 2 and 3 only are correct

(d)- 1 only is correct

Answer
Verified

1 mol of an ideal gas has the volume at STP = 22.4 L.

Hence, \[{{V}_{m}}=\text{ }22.4L\]

So, the number of moles is equal to the given volume upon volume of 1 mol gas.

We have to find the number of atoms of carbon and nitrogen in C H N.

So, we can account x to C, y to H, and z to N.

$({{C}_{x}}{{H}_{y}}{{N}_{z}})$

Therefore, number of moles of the compounds will be:

$n({{C}_{x}}{{H}_{y}}{{N}_{z}})=\dfrac{V({{C}_{x}}{{H}_{y}}{{N}_{z}})}{{{V}_{m}}}=\dfrac{10c{{m}^{3}}}{{{V}_{m}}}$

$n({{N}_{2}})=\dfrac{V({{N}_{2}})}{{{V}_{m}}}=\dfrac{5c{{m}^{3}}}{{{V}_{m}}}$

$n(C{{O}_{2}})=\dfrac{V(C{{O}_{2}})}{{{V}_{m}}}=\dfrac{20c{{m}^{3}}}{{{V}_{m}}}$

According to the question, the schematic equation of the chemical reaction will be:

${{C}_{x}}{{H}_{y}}{{N}_{z}}\text{ }+\text{ }(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O+\dfrac{z}{2}{{N}_{2}}$

So, by formulating with the help of the above reaction we can find the value of x and y.

We can write,

$n({{C}_{x}}{{H}_{y}}{{N}_{z}})=\dfrac{n(C{{O}_{2}})}{x}=\dfrac{2\text{ x }n({{N}_{2}})}{z}$

For, x

$n({{C}_{x}}{{H}_{y}}{{N}_{z}})=\dfrac{n(C{{O}_{2}})}{x}$

$x=\dfrac{n(C{{O}_{2}})}{n({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\frac{V(C{{O}_{2}})}{V({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\dfrac{20c{{m}^{3}}}{10c{{m}^{3}}}=2$

So, the value of x is 2

For, z

$\dfrac{z}{2}=\dfrac{n({{N}_{2}})}{n({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\dfrac{V({{N}_{2}})}{V({{C}_{x}}{{H}_{y}}{{N}_{z}})}=\dfrac{5c{{m}^{3}}}{10c{{m}^{3}}}=\dfrac{1}{2}$

$z=\dfrac{1}{2}\text{ x 2 = 1}$

So, the value of z is 1.

So, in the J compound, the carbon atoms are 2, and the number of nitrogen atoms is 1.

Hence, the correct option is (c)- 2, and 3 only are correct.