
Given, KF Has NaCl type structure. What is the distance (in A˚) between ${K^ + }$ and ${F^ - }$ in $KF$ if the density of $KF$ is $2.48gc{m^{ - 3}}$?
A) $2.685$
B) $4.632$
C) $6.235$
D) None of the above
Answer
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Hint: As we know that the potassium fluoride is the substance compound with the equation \[KF\]. After hydrogen fluoride, \[KF\] is the essential wellspring of the fluoride particle for applications in assembling and in science. It's anything but an antacid halide and happens normally as the uncommon mineral carobbiite. Solution of \[KF\] will scratch glass because of the development of dissolvable fluorosilicates, despite the fact that \[HF\] is more successful.
Complete answer:
The density of any substance can be calculated using the formula,
$d = \dfrac{{n\left( M \right)}}{{V\left( {{N_a}} \right)}}$
The structure of $NaCl$ is framed by rehashing the face focused cubic unit cell. It has \[1:1\] stoichiometry proportion of \[Na:Cl\] with a molar mass of \[58.4g/mol\]
It is given that the density of potassium fluoride is $2.48gc{m^{ - 3}}$.
By substituting the know value in the above formula,
$a = 5.37{A^0}$
$2.48 = \dfrac{{4\left( {58} \right)}}{{{a^3}\left( {6.022 \times {{10}^{23}}} \right)}}$
On simplification we get,
$a = 5.37 \times {10^8}cm$
The distance between potassium ion and fluoride ion is given as,
$Dis\tan ce = \dfrac{a}{2} = 2.685{A^0}$
So, the correct answer is “Option A”.
Note:
Sodium chloride, generally known as salt (in spite of the fact that ocean salt additionally contains other substance salts), is an ionic compound addressing a \[1:1\] proportion of sodium and chloride particles. With molar masses of \[22.99\] and \[35.45g/mol\] separately, \[100g\] of \[NaCl\] contains \[39.34g\] $Na$ and \[60.66g\] $Cl$ . Sodium chloride is the salt generally liable for the saltiness of seawater and of the extracellular liquid of numerous multicellular life forms. In its palatable type of table salt, it is normally utilized as a topping and food additive. Huge amounts of sodium chloride are utilized in numerous modern cycles, and it's anything but a significant wellspring of sodium and chlorine compounds utilized as feedstocks for additional synthetic amalgamations. A second significant use of sodium chloride is de-icing of streets in sub-frigid atmosphere.
Complete answer:
The density of any substance can be calculated using the formula,
$d = \dfrac{{n\left( M \right)}}{{V\left( {{N_a}} \right)}}$
The structure of $NaCl$ is framed by rehashing the face focused cubic unit cell. It has \[1:1\] stoichiometry proportion of \[Na:Cl\] with a molar mass of \[58.4g/mol\]
It is given that the density of potassium fluoride is $2.48gc{m^{ - 3}}$.
By substituting the know value in the above formula,
$a = 5.37{A^0}$
$2.48 = \dfrac{{4\left( {58} \right)}}{{{a^3}\left( {6.022 \times {{10}^{23}}} \right)}}$
On simplification we get,
$a = 5.37 \times {10^8}cm$
The distance between potassium ion and fluoride ion is given as,
$Dis\tan ce = \dfrac{a}{2} = 2.685{A^0}$
So, the correct answer is “Option A”.
Note:
Sodium chloride, generally known as salt (in spite of the fact that ocean salt additionally contains other substance salts), is an ionic compound addressing a \[1:1\] proportion of sodium and chloride particles. With molar masses of \[22.99\] and \[35.45g/mol\] separately, \[100g\] of \[NaCl\] contains \[39.34g\] $Na$ and \[60.66g\] $Cl$ . Sodium chloride is the salt generally liable for the saltiness of seawater and of the extracellular liquid of numerous multicellular life forms. In its palatable type of table salt, it is normally utilized as a topping and food additive. Huge amounts of sodium chloride are utilized in numerous modern cycles, and it's anything but a significant wellspring of sodium and chlorine compounds utilized as feedstocks for additional synthetic amalgamations. A second significant use of sodium chloride is de-icing of streets in sub-frigid atmosphere.
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