
The vapor density of fully dissociated $N{H_4}Cl$ would be:
A. Less than half of the vapor density of pure $N{H_4}Cl$ .
B. Double of the vapor density of pure $N{H_4}Cl$ .
C. Half of the vapor density of pure $N{H_4}Cl$ .
D. One-third of the vapor density of pure $N{H_4}Cl$ .
Answer
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Hint: Vapor density is defined as the density of vapor of a given gas to that of the density of hydrogen gas. It is also defined as the half of the molecular weight of the given gaseous compound. It can be also defined as the ratio of molar mass of a given gas to that of the molar mass of hydrogen gas.
Complete step by step answer:
The complete dissociation of ammonium chloride produces ammonia gas and hydrochloric acid. The reaction is as follows:
$N{H_4}Cl \to N{H_3} + HCl$
As we know that the molecular weight of the compound is twice of that of its vapor density which can be shown mathematically as:
${M_w} = 2 \times V.D$
Let us determine the molecular weight of ammonium chloride ($N{H_4}Cl$ ), ammonia ($N{H_3}$ ) and hydrochloric acid ($HCl$ ).
$N{H_4}Cl = 14 + 4 \times 1 + 35.5 = 53.5g/mol$
$N{H_3} = 14 + 3 \times 1 = 17g/mol$
$HCl = 1 + 35.5 = 36.5g/mol$
Thus, the average molecular weight of any compound is determined by:
$AAW = \dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{{n_1} + {n_2}}}$
Where, ${n_1} = $ Stoichiometric coefficient of first product
${M_1} = $ Molecular weight of first product
${n_2} = $ Stoichiometric coefficient of second product
${M_2} = $ Molecular weight of second product
$AAW = $ Average atomic weight
Now, substituting the values, we get:
$AAW = \dfrac{{1 \times 17 + 1 \times 36.5}}{{1 + 1}}$
On solving, we have:
$AAW = 26.75g/mol$
As we can see that the molecular weight of ammonium chloride is double the average atomic weight of the products of the dissociation.
Thus, the correct option is C. Half of the vapor density of pure $N{H_4}Cl$ .
Note:
The application of the vapor density is that it indicates whether a gas is denser (greater than one) or less dense (less than one) than air. The density has its applications in container storage and personnel safety. Thus, if a container releases a dense gas, its vapor can sink down and if is flammable, one can collect until it attains a concentration sufficient for ignition.
Complete step by step answer:
The complete dissociation of ammonium chloride produces ammonia gas and hydrochloric acid. The reaction is as follows:
$N{H_4}Cl \to N{H_3} + HCl$
As we know that the molecular weight of the compound is twice of that of its vapor density which can be shown mathematically as:
${M_w} = 2 \times V.D$
Let us determine the molecular weight of ammonium chloride ($N{H_4}Cl$ ), ammonia ($N{H_3}$ ) and hydrochloric acid ($HCl$ ).
$N{H_4}Cl = 14 + 4 \times 1 + 35.5 = 53.5g/mol$
$N{H_3} = 14 + 3 \times 1 = 17g/mol$
$HCl = 1 + 35.5 = 36.5g/mol$
Thus, the average molecular weight of any compound is determined by:
$AAW = \dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{{n_1} + {n_2}}}$
Where, ${n_1} = $ Stoichiometric coefficient of first product
${M_1} = $ Molecular weight of first product
${n_2} = $ Stoichiometric coefficient of second product
${M_2} = $ Molecular weight of second product
$AAW = $ Average atomic weight
Now, substituting the values, we get:
$AAW = \dfrac{{1 \times 17 + 1 \times 36.5}}{{1 + 1}}$
On solving, we have:
$AAW = 26.75g/mol$
As we can see that the molecular weight of ammonium chloride is double the average atomic weight of the products of the dissociation.
Thus, the correct option is C. Half of the vapor density of pure $N{H_4}Cl$ .
Note:
The application of the vapor density is that it indicates whether a gas is denser (greater than one) or less dense (less than one) than air. The density has its applications in container storage and personnel safety. Thus, if a container releases a dense gas, its vapor can sink down and if is flammable, one can collect until it attains a concentration sufficient for ignition.
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