Solid State Test

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Solid State Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

A compound $$AB$$ has rock salt type structure. The formula weight of $$AB$$ is $$6.023 \ Y$$ amu, and closest $$A-B$$ distance is $$Y^{1/3}$$ nm, where $$Y$$ is an arbitrary number, find the density of the lattice.

A
$$5$$ kg$$/m^3$$

B
$$5\times 10^{-3}$$ g$$/m^3$$

C
$$10^{-3}$$ g$$/m^3$$

D
$$15\times 10^{-2}$$ g$$/m^3$$

Solution

Let $$AB$$ has rock salt structure $$(A:B::1:1)$$
For $$F.C.C$$ structure $$(n=4)$$ and formula weight of $$AB$$ is $$6.023 y g
Closest distance $$A - B = y\frac{1}{3}nm$$
Therefore,
Edge length of unit cell,
$$a = 2\left( {{A^ + }{B^ - }} \right) = 2 \times \frac{y}{3} \times {10^{ - 9}}m$$
Therefore ,
Density of $$AB = \frac{{n \times mol.wt}}{{{N_A}V}}$$
$$ = \frac{{4 \times 6.023 \times y \times {{10}^{ - 3}}}}{{6.023 \times {{10}^{23}} \times {{\left( {2 \times \frac{y}{3}{{10}^{ - 9}}} \right)}^3}}}$$
$$ = 5.0kg\,{m^{ - 3}}$$.
Question 2
1 / -0

Potassium has a bcc structure wit nearest neighbour distance $$4.52A^0$$, its atomic weight is $$39$$. Its density (in kg $$m^{-1}$$) will be:

A
$$454$$

B
$$804$$

C
$$852$$

D
$$908$$

Solution

For b.c.c, $$d = \dfrac{{\sqrt 3 a}}{2}$$
or $$a = \dfrac{{2d}}{{\sqrt 3 }}$$
$$a=\dfrac{{2 \times 4.52}}{{1.732}}$$
$$a=5.219{A^ \circ } = 522\,pm$$
Hence,
$$\rho = \dfrac{{z \times M}}{{{a^3} \times {N_A} \times {{10}^{ - 30}}}}$$
Implies that
$$\rho=\dfrac{{2 \times 39}}{{{{(522)}^3} \times 6.023 \times {{10}^{23}} \times {{10}^{ - 30}}}}$$
$$\rho=0.91g/c{m^3}$$
$$\rho=910kg{m^{ - 3}}$$
Question 3
1 / -0

The maximum radius of sphere that can be fitted in the octahedral hole of cubical closed packing of sphere of radius $$r$$ is:

A
$$0.732\ r$$

B
$$0.225\ r$$

C
$$0.155\ r$$

D
$$0.414\ r$$

Solution

For octahedral voids, $$r+r_1=2r\cdot \cfrac{1}{\sqrt{2}}$$
$$\implies \cfrac{r_1}{r}=\sqrt 2 r$$
$$\implies \cfrac{r_1}{r}=\sqrt{2}-1=0.414$$
$$\implies r_1=0.414r$$
Question 4
1 / -0

A solid $$X^+Y^-$$ has a bcc structure. If the distance of closest approach between the two atoms is $$173$$ pm, the edge length of the cell is:

A
$$200$$ pm

B
$$\sqrt{3}/\sqrt{2}$$ pm

C
$$142.2$$ pm

D
$$\sqrt{2}$$ pm

Solution

For BCC lattice $$r=0.433a$$
Lattice is half of the body diagonal , that is $$\dfrac{\sqrt{3}a}{2}=173\ pm$$
Thus
$$a=\dfrac{173\times2}{\sqrt{3}}$$
$$=\dfrac{173\times2}{1.73}$$
$$=200\ pm$$
Question 5
1 / -0

The unit cell length of $$NaCl$$ is observed to be $$0.5627$$ nm by X-ray difference studies; the measured density of $$NaCl$$ is $$2.164$$g $$cm^{-3}$$/ Calculate the difference of observed and calculated density. lso calculate $$\%$$ is missing $$Na^+$$ and $$Cl^-$$ ions:

A
$$0.031, 0.775\%$$

B
$$3.96, 0.031\%$$

C
$$0.0031, 7.75\%$$

D
$$1.7, 1.75\%$$

Solution

$$\rho=\cfrac{M_A\times Z}{N_A\times a^3}$$
$$\rho_{observed}=2.164$$ $$gcm^{-3}$$
$$a=0.5627$$ $$nm$$
$$N_A=6.023\times 10^{23}$$
$$Z_{observed}=?$$
$$2.164=\cfrac{58.5\times Z_{observed}}{6.023\times (5.627)^3\times 10^{-24}\times 10^{23}}$$
$$Z_{observed}=3.96$$
$$Z_{theoretical}=4$$
$$\rho_{theoretical}=2.180$$ $${g/cm^3}$$
So, $$\rho=\cfrac{4\times M_A}{N_A\times a^3}$$
$$\rho_{theoretical}-\rho_{observed}=0.016$$ $${gm/cm^3}$$
Percentage of missing ions $$=\cfrac{4-396}{4}\times 100=1\%$$
Question 6
1 / -0

The number of unit cells in $$58.5$$ g of $$NaCl$$ is nearly ?

A
$$6 \times 10^{23}$$

B
$$3\times 10^{22}$$

C
$$1.5\times 10^{23}$$

D
$$0.5\times 10^{24}$$

Solution

$$NaCl$$ crystalline in FCC structure i.e., $$4$$ molecules of $$NaCl$$ is present in one unit cell.
Moles of $$NaCl=\cfrac{58.5}{58.5}=1$$
Moles of unit cell $$=\cfrac{1}{4}=0.25$$
Total number of unit cells $$=\cfrac{1}{4}\times 6\times 10^{23}=1.5\times 10^{23}$$ cells
Question 7
1 / -0

A crystalline solid $$AB$$ adopts sodium chloride type structure with edge length of the unit cell as 745 pm and formula mass of 74.5 a.m.u. The density of the crystalline compound is:

A
$$2.16 g cm^{-3}$$

B
$$0.99g cm^{-3}$$

C
$$1.88g cm^{-3}$$

D
$$1.97 g cm^{-3}$$

Solution

Density $$=\rho=\cfrac{Z\times M_A}{N_A\times a^3}$$
$$N_A=6.023\times 10^{23}$$
$$Z_{NaCl}=4$$ (FCC structure)
$$M_{AB}=74.5$$ $$amu$$
$$a=7.45\mathring A$$
$$\rho=\cfrac{4\times 74.5}{6.023\times 10^{23}\times (7.45)^3\times 10^{-24}}=1.197$$ $${g/cm^3}$$
Question 8
1 / -0

The fraction of octahedral voids filled by $$A{l^{3 + }}$$ ions in $$A{l_2}{O_3}\left( {{r_{Al^{3 \oplus }}}}/r_{O^{2 - }} = 0.43 \right)$$ is:

A
0.43

B
0.287

C
0.667

D
1

Solution

For $$A l_{2} O_{3}, \dfrac{\gamma_{A L}+3}{\gamma_{0}^{2}-}=0.43$$.
$$ \begin{array}{l} \text { It forms hexagonal closed packed lattice in which } A l^{+3} \text {ions } \\ \text { occupy } \dfrac{2}{3} \text { rd of } \text { octahedral voids. } \\ \text { so, option (c) is correct. } \end{array} $$
Question 9
1 / -0

Polonium crystallizes in a simple cubic structure. The edge of the unit cell is $$0.236$$ nm. What is the radius of the polonium atom?

A
$$0.144$$ nm

B
$$0.156$$ nm

C
$$0.118$$ nm

D
$$0.102$$ nm

Solution

For simple cubic structure
Edgelength $$=a=2r$$
$$a=0.236$$ $$nm$$
$$r=\cfrac{a}{2}=0.118$$ $$nm$$
Question 10
1 / -0

In the crystal $$A^{2+}B^{2-}$$, having anions in the face centred cubic packing if the radius of the anion if $$1.84A^0$$, ideal radius of the cation present in the tetrahedral hole will be:

A
$$0.225A^0$$

B
$$0.414A^0$$

C
$$0.732A^0$$

D
none of these

Solution

For tetrahedral holes, ideal ratio of $$\cfrac{r_+}{r_-}$$ is $$0.225$$
Given: $$r_-=1.84\mathring A$$
Then,
$$\cfrac{r_+}{1.84}=0.225$$
$$r_+=0.414 \mathring A$$

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Re-Attempt Weekly Quiz Competition

Solid State Test - 44

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Solid State Test - 44

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Question 1

$$5$$ kg$$/m^3$$

$$5\times 10^{-3}$$ g$$/m^3$$

$$10^{-3}$$ g$$/m^3$$

$$15\times 10^{-2}$$ g$$/m^3$$

Solution

Question 2

$$454$$

$$804$$

$$852$$

$$908$$

Solution

Question 3

$$0.732\ r$$

$$0.225\ r$$

$$0.155\ r$$

$$0.414\ r$$

Solution

Question 4

$$200$$ pm

$$\sqrt{3}/\sqrt{2}$$ pm

$$142.2$$ pm

$$\sqrt{2}$$ pm

Solution

Question 5

$$0.031, 0.775\%$$

$$3.96, 0.031\%$$

$$0.0031, 7.75\%$$

$$1.7, 1.75\%$$

Solution

Question 6

$$6 \times 10^{23}$$

$$3\times 10^{22}$$

$$1.5\times 10^{23}$$

$$0.5\times 10^{24}$$

Solution

Question 7

$$2.16 g cm^{-3}$$

$$0.99g cm^{-3}$$

$$1.88g cm^{-3}$$

$$1.97 g cm^{-3}$$

Solution

Question 8

0.43

0.287

0.667

1

Solution

Question 9

$$0.144$$ nm

$$0.156$$ nm

$$0.118$$ nm

$$0.102$$ nm

Solution

Question 10

$$0.225A^0$$

$$0.414A^0$$

$$0.732A^0$$

none of these

Solution

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