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Solid State Test - 34

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Solid State Test - 34
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  • Question 1
    1 / -0
    A gaseous mixture contains 56 g of $$N_2$$ , 44 g of $$CO_2$$ and 16 g of $$CH_4$$. The total pressure of the mixture is 720 mm of Hg. The partial pressure of methane is
    Solution

  • Question 2
    1 / -0
    What type of bonding results in compounds which are crystalline solids?
    Solution
    Ionic bonding results in compounds which are crystalline solids. As they are held by strong electrostatic forces, thus they have more organised structure, a definite melting point and are known as crystals.
  • Question 3
    1 / -0
    Glass is an example of this type of solid.
    Solution
    In the glass, the constituents particles are arranged in a random manner.
    Hence it is an Amorphous solid.
    Glass exhibits the properties of an amorphous solid like:
    1. They do not have sharp melting and boiling point.
    2. They have isotropic, etc.
  • Question 4
    1 / -0
    The following image is a cross section of a solid crystal of Iodine $$(I_{2})$$.
    What type of crystal is represented by this Iodine solid?

    Solution
    Here, Iodine crystallizes into ionic type crystal because the given diagram is of simple cubic unit cell which are very orderly arranged i.e. Iodine atoms at body corners.
  • Question 5
    1 / -0
    A certain gas takes three times as long to effuse out as helium. Its molecular mass will be :
    Solution
    $$\begin{array}{l}\text { We know, rate of effusion (r) } \alpha \text { } \dfrac{1}{\sqrt{M}} \text { , where } M \text { is the molecular }\\\text {weight of gas. Let, the molecular mass of required gas be M. }\end{array}$$

    $$\begin{array}{l}\text { Then, } \dfrac{r_{m}}{r_{H e}}=\dfrac{1}{3}\\\Rightarrow \sqrt{\dfrac{4}{M}}=\dfrac{1}{3} \Rightarrow \dfrac{4}{M}=\dfrac{1}{9} \Rightarrow M=36 \\\text { Thus, molecular mass of required } gas=36u\end{array}$$
    Option B is correct answer
  • Question 6
    1 / -0
    Which of the following have a regular, repeated molecular pattern and large number of free surfaces in three dimensional space?
    Solution
    As we know gases do not form surfaces and liquids does not have free space. so the answer is solid and regular repeated pattern present in solids.
  • Question 7
    1 / -0
    The packing efficiency of simple cubic $$(sc)$$, body centered cubic $$(bcc)$$ and cubic close packing $$(ccp)$$ lattices follows the order:
    Solution
    The packing efficiency of simple cubic(SC), body centered cubic (bcc) and cubic close packing (ccp) lattices follows the order :
    scc (52.4 %) < bcc (68 %) < ccp (74 %).
    SC is loosely bonded. Almost half the space is empty. CCP is the most efficient packing.
    Option (D) is the correct answer.
  • Question 8
    1 / -0

    The ratio of cations to anion in a closed pack tetrahedral is:

    Solution
    Because there is only one atom per unit cell, and an atom has a volume equal to $$\dfrac{4}{3}(πr^3)$$, we can also determine the packing efficiency: Simple cubic structures have a packing efficiency of 52%.
    $$\dfrac{r^+}{r^-}$$ for tetrahedral void = 0.225 - 0.414
    $$\dfrac{r^+}{r^-}$$ for triangular void = 0.155 - 0.225
  • Question 9
    1 / -0
    Which of the following arrangements correctly represents hexagonal and cubic close packed structure respectively?
    Solution
    $$ABAB.......$$ and $$ACBACB........$$  arrangements correctly represents hexagonal and cubic close-packed structure respectively.
    Note: $$AAAA$$ type arrangement represents simple cubic structure.
    $$ACBACB........$$  arrangement can also be represented as $$ABCABC........$$  arrangement.
  • Question 10
    1 / -0
    The name of the $$NaCl$$ crystal unit cell and its packing efficiency are respectively:
    Solution
    $$NaCl$$ has $$FCC$$ crystal unit cell.
    $$FCC$$(Face Centered Cubic) lattice each ion is $$6-$$ coordinate and has a local octahedral geometry.
    $$FCC$$ unit cell contains $$4$$ number of atoms.
    In $$\triangle ABC$$,
    $$AC^{2}=AB^{2}+BC^{2}=b^{2}$$
    $$\Rightarrow a^{2}+a^{2}=b^{2}$$
    $$\Rightarrow 2a^{2}=(4\pi )^{2}$$
    $$\Rightarrow a=\sqrt{\cfrac{16}{2}}r$$
    $$\Rightarrow a=\cfrac{4}{\sqrt{2}}r=2\sqrt{2}r$$
    Volume of one sphere$$=\cfrac{4}{3}\pi r^{3}$$
    $$\because 4$$ atoms , so $$4$$ spheres
    Volume of $$4$$ sphere $$=4\times \cfrac{4}{3}\pi r^{3}$$
    and volume of cube $$=a^{3}= (2\sqrt{2}r)^{3}$$
    Therefore, $$Packing\quad efficiency=\cfrac{Volume\quad occupied\quad by\quad 4\quad sphere\quad in\quad unit\quad cell\ast 100 }{Volume\quad of\quad unit\quad cell}$$
    $$\Rightarrow Packing\quad efficiency= \cfrac{4\times \cfrac{4}{3}\pi r^{3}}{(2\sqrt{2})^{3}(r)^{3}} \times 100= 74\%$$

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