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mr Goodwill [35]
3 years ago
15

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 417 pm. The diameter of the metal atom is:

_______
a. 417 pm
b. 295 pm
c. 209 pm
d. 147 pm
e. none of these
Chemistry
1 answer:
enyata [817]3 years ago
8 0

Answer:

b. 295 pm

Explanation:

To answer this question we need to use the equation of a face-centered cubic laticce:

Edge length = √8 R

<em>Where R is radius of the atom.</em>

<em />

Replacing:

417pm = √8 R

R = 147.4pm is the radius of the atom

As diameter = 2 radius.

Diameter of the metal atom is:

147.4pm* 2 =

295pm

Right answer is:

<h3>b. 295 pm </h3>

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Explanation:

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Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?
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Answer : The correct option is, (A) AlCl_3

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

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It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of AlCl_3 will be,

AlCl_3\rightarrow Al^{3+}+3Cl^{-}

So, Van't Hoff factor = Number of solute particles = Al^{3+}+3Cl^{-} = 1 + 3 = 4

(B) KI

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The dissociation of KI will be,

KI\rightarrow K^{+}+I^{-}

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(C) CaCl_2

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The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

So, Van't Hoff factor = Number of solute particles = Ca^{2+}+2Cl^{-} = 1 + 2 = 3

(D) MgSO_4

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of MgSO_4 will be,

MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}

So, Van't Hoff factor = Number of solute particles = Mg^{2+}+SO_4^{2-} = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, AlCl_3

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