Solid state | Practice questions | CBSE 12

1. If  Radius (r) = 125 pm and  z = 4 (FCC),find the Edge length of unit cell (a) and Number of unit cells in 1 cm3 of Al = ?  

      Answer: 2.266 × 1024 unit cells

2. If Edge length (a) = 3.9231 Å or 3.9231 × 10−8 cm,  Atomic weight of Pt (M)= 195.08  and 

      z = 4 (FCC) find the Density (d) and Atomic radius (r) = ? 

      Avogadro’s number = NA = 6.023 × 1023 

      Answer:  r = 138 .7 pm

3. Edge length (a) = 4.049 Å, Atomic mass of Al (M) = 26.98 g/mol and z = 4 (FCC) find Density (d) = ? 

     Avogadro’s number = NA = 6.023 × 1023

     Answer: 2.699 g/cc

4. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and that of P are at the bodycentre. What is the formula of the compound? What are the coordination numbers of P and Q? 

     Answer: The formula of the compound is PQ.The coordination number of both P and Q is 8.

5. The edge length of the unit cell of Ta, is 330.6 pm; the unit cell is body-centered cubic. Tantalum has a density of 16.69 g/cm3 (a) calculate the mass of a tantalum atom. (b) Calculate the atomic weight of tantalum in g/mol.

      Answer: 3.015 x 10¯22 g and 181.6 g/mol 

6. An element with density 11.2 g cm-3 forms an FCC lattice with edge length of 4 X 10-8cm. Calculate the atomic mass of the element. (Given: N = 6.022 X 1023 mol-1)

        Answer: 107.9 g 

7. Tungsten crystallizes in a body-centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of the tungsten atom?

       Answer: 137.04 pm

8. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca2+ ions and 8 F–. ions and that Ca2+ ions are arranged in a FCC lattice. The F- ions fill all tetrahedral holes in the face centred cubic lattice of Ca2+ ions. The edge of the unit cell is 5.46 X 10-8 cm in length. The density of the solid is 3.18 g cm-3. Use this information to calculate Avogadro’s number. (Molar mass of CaF2 = 78.08 g mol-1)

       Answer: 6.023 X 10 23 

9. What is the formula of a compound in which the element ‘Y’ form hep lattice and atoms of ‘X’ occupy 1/3 r“ of octahedral voids.

             Answer: XY3 

10. What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 2/3rd of tetrahedral voids?

Answer: X4 Y1

11. Calculate the packing efficiency of a metal crystal for simple cubic lattice. 

              Answer: 52.4%

12. Why does Li Cl acquire pink colour when heated in Li vapours?

13. What type of semiconductor is obtained when silicon is doped with boron?

14. Express the relationship between atomic radius(r) and the edge length (a) of the FCC unit cell.

15. What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy l/3rd of tetrahedral voids? 

            Answer : X ⅔ : Y 

Mole Concept | Limiting reagent | Stoichiometry | Practice questions

QUESTION 1:

In this reaction:

CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O

6.088 g CaCO3 reacted with 2.852 g HCl. What mass of CaCO3 remains unreacted?

                                                                                                                                        

SOLUTION :

HCl is Limiting reagent

CaCO3 ⇒ 6.088 g / 100.086 g/mol = 0.0608277 mol

HCl ⇒ 2.852 g / 36.461 g/mol = 0.0782206 mol

Grams of calcium carbonate used:

1 is to 2 as x is to 0.0782206 mol

x = 0.0391103 mol

0.0391103 mol times 100.086 g/mol = 3.914 g

Grams of CaCO3 remaining: 6.088 g minus 3.914 g = 2.174 g

QUESTION 2:

950.0 grams of copper(II) sulfate are reacted with 460.0 grams of zinc metal. (a) What is the theoretical yield of Cu? (b) If 295.8 grams of copper are actually obtained from this reaction, what is the percent yield?

SOLUTION: 

1. The balanced chemical equation: CuSO4 + Zn ---> ZnSO4 + Cu

According to the balanced chemical equation, calculate the moles of reactants. 

CuSO4 ⇒ 950.0 g / 159.607 g/mol = 5.95212 mol

Zn ⇒ 460.0 g / 65.38 g/mol = 7.03579 mol 

So, CuSO4 is limiting reagent. 

Grams of Cu: 5.95212 mol of Cu is produced (as 1 : 1 molar ratio involved)

5.95212 mol X  63.546 g/mol = 378.2 g

2. Percent yield : 295.8 g / 378.2 g = 78.21 %

QUESTION 3:

Silicon nitride (Si3N4) is made by combining Si and nitrogen gas (N2) at a high temperature. How much (in g) Si is needed to react with an excess of nitrogen gas to prepare 125 g of silicon nitride if the percent yield of the reaction is 95.0%?

Answer: 79.1 g Si

QUESTION 4:

Use the following reaction: C4H9OH + NaBr + H2SO4 C4H9Br + NaHSO4 + H2O

If 15.0 g of C4H9OH reacts with 22.4 g of NaBr and 32.7 g of H2SO4 to yield 17.1 g of C4H9Br, what is the percent yield of this reaction?

Answer: 61.5 %.

QUESTION 5:

If 4.95 g of ethylene (C2H4) are combusted with 3.25 g of oxygen.

a. What is the limiting reagent?

b. How many grams of CO2 are formed?

Answer: Oxygen and 2.98 g CO2

QUESTION 6:

If you dissolve lead(II) nitrate and potassium iodide in water they will react to form lead(II) iodide and potassium nitrate. 

a.Write the equation for this reaction, then balance the equation. 

b.Calculate the grams of lead(II) iodide that can be produced from 5.00 moles of potassium iodide. 

c.Calculate the grams of lead(II) iodide that can be produced from 75.00 grams of potassium iodide.

Answer:  1.15 x 103 g PbI2 and  104.1 g PbI2 

QUESTION 7:

Write then balance the combustion reaction for propane gas, C3H8. 

a. If 5.00 grams of propane burn completely, what volume of carbon dioxide is produced at STP? 

b. If 75.0 L of steam is produced at STP, what mass of propane must have burned? 

c. If 34.2 grams of propane are completely combusted, how many moles of steam will that produce?

Answer: 7.62 L CO2  ,  36.9 g C3H8 and  3.10 mol steam

QUESTION 8:

If 31.6 grams of magnesium hydroxide is combined with 68.0 mL of 0.725 M HBr, which is the limiting reagent? How many grams of magnesium bromide would be formed? 

Answer: HBr is the limiting reagent and 4.54 g MgBr2

QUESTION 9:

Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains?

The unbalanced equation is:

Al2S3 + H2O ---> Al(OH)3 + H2S

Answer: Al2S3 is the limiting reagent.227.4685 g of water and 265.5 g excess.

QUESTION 10:

What weight of each substance is present after 0.4500 g of P4O10 and 1.5000 g of PCl5 are reacted completely?

P4O10 + 6PCl5 ---> 10POCl3

Answer: PCl5 is limiting, 0.1092 g of P4O10 remaining,  1.8408 g of  POCl3 produced

Solutions | Raoult's law | Solved problems

QUESTION 1:

Calculate the osmotic pressure of a solution 0.1 molar of sucrose (Mw = 342) at  20oC.

SOLUTION:  

From the formulae,  

Molality = moles/kilogram of solvent

Molarity = moles/liter of solution

The grams of sucrose in the solutions are 34.2 and the total weight of the solution is

34.2 + 1000 grams of solvent = 1034.2

Assuming a density is 1 

          0.1 moles : 1034.2 = M : 1000 grams of solution

          M = 100/1034.2 = 0.097

          p = (0.097)(0.0820578 liter atm mol-1 K-1)(293) = 2.34 atm

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QUESTION 2:

A mixture of water and acetone at 756 mm boils at 70oC. Calculate the percentage composition of the  mixture using the following table:

Temperature oC	Vapor pressure (atm) Acetone	Vapor pressure (atm) Water
60	1.14	0.198
70	1.58	0.312
80	2.12	0.456
90	2.81	0.694

SOLUTION:  

According to Raoult's law, 

P = xacetone Po acetone + xwater Powater 

    = 756/760 =0.995 atm

Substituting the values at 70oC ,  

P = xacetone 1.58 + xacqua 0.312 = 0.995

We also know that, 

xacetone + xacqua = 1

So, xacqua = 1 - xacetone

  So,  xacetone 1.58 + 0.312 (1 - xacetone) = 0.995

xacqua 1.58 + 0.312 - 0.312 xacetone = 0.995

xacetone 1.26+ 0.312 = 0.995

xacetone = (0.995 - 0.312) / 1.26 = 0.54

So,  xwater = 1 - 0.54 = 0.46

Considering that we have 100 moles of solution, 54 moles of  acetone and 46 moles of water is present in the sample. 

54 moles of  acetone is 54 x 58 =3132 grams

and 46 moles of water is 46 x 18 = 828 grams

So,  % water = (828)(100)/3132 + 828) = 21%

       % acetone = 100-21 = 79%

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QUESTION 3:

The addition of 114 grams of sucrose to 1000 grams of water lowers the vapor pressure of water from 17.540 to 17.435. Calculate the molecular weight of sucrose.

SOLUTION:  

The difference in pressure: 17.540 - 17.435 = 0.105 

Then, 

0.105 = 17.540 x2  = nsolute/(nsolvent + nsolute)

nsolvent = 1000/mw water = 1000/18 = 55.55 moles

0.105 = 17.540 x2  = 17.540 nsolute/(55.55 + nsolute)

by solving this,

         nsolute = 0.335 moles and 

MW = 114/0.335 = 340

The calculated molecular weight (340) is very close to the true molecular weight of sucrose 342

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QUESTION 4:

Calculate the osmotic pressure at 20o of a suspension containing 60g/l of solid particles each particle having a mass of 10-9 grams (1 nanogram).

SOLUTION:   

Formula:  p = MRT ( T = 278 + 20 = 298 oK and R = 0.0823)

The number of particles in one liter of solution is = 60/10-9 = 6 x 1010

1 mole of any substance contains an Avogadro number of particles (6 x 1023)

The number of moles in one liter of suspension = 6 x 1010/6 x 1023 =  10-13

which is equal to the molarity of the suspension.

So, p = (0.0823)(298) (10-13) = 24.5 x 10-13 atm

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QUESTION 5:

Calculate the vapor pressure of a mixture containing  252 g of n-pentane (Mw = 72) and 1400 g of n-heptane (Mw = 100) at 20oC. The vapor pressure of n-pentane and n-heptane are 420 mm Hg and 36 mm Hg respectively.

SOLUTION: 

From  Raoult's law, P = Po x

Here,

P = vapor pressure of the component in the mixture.

Po= vapor pressure of the pure component.

x = Molar fraction of the component in the mixture.

Calculation of mole fraction: 

Mole fraction = moles of one component / Total moles in solution

Moles n-pentane = 252/72 = 3.5

Moles n-eptano = 1400/100 = 14

Totals = 3.5 + 14 = 17.5 moles

xn-pentane = 3.5/17.5 = 0.2 ,  xn-eptane = 14/17.5 = 0.8

So,  Pn-pentane =  0.2 x 420 = 84 mm Hg ; Pn-eptane =  0.8 x 36 = 28.8 mm Hg

and the vapor pressure of mixture is, Pmixture = 84 + 28.8 = 112.8 mm

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QUESTION 6:

Calculate the boiling point (at 1 atm) of a solution containing 116 g of acetone (Mw = 58) and 72 g of water (Mw = 18) by using the following table:

Temperature oC	Vapor pressure (atm) Acetone	Vapor pressure (atm) Water
60	1.14	0.198
70	1.58	0.312
80	2.12	0.456
90	2.81	0.694

SOLUTION: 

A liquid starts to boil when its vapor pressure matches the atmospheric pressure (1 atm in this case).

According to Raoult's law

P = xacetone Po acetone + xwater Powater = 1 atm

Calculate the molar fractions: 

Mole fraction = moles of one component / Total moles in solution

molesacetone = 116/58 = 2 ; moleswater = 72/18 = 4

So, Total moles = 2 + 4 =  6 

Substituting the given values, 

xacetone = 2/4 = 1/3  ;  xwater = 4/6 = 2/3

So,  P = 2/3 Po acetone + 1/3 Powater= 1 atm

By trials, using the table, we can find the values of vapor pressure which satisfies the above equation. The best result is obtained by using  the values at  80oC :

P = 2/3 0.456 + 1/3 2.12 = 1.01 atm

So, the boiling point is about 80oC.