FSc 1 Chemistry | Basic Concepts | Short Question Answers - 1

 

FSc Part-I Chemistry | Basic Concepts | Short Question Answers - 1. The following short questions and their answers from Chapter No.1: Basic Concepts are the most important for the final examinations:

Q.1: Define molecule, give examples. (Lahore Board 2013)

Answer: 

A molecule is the smallest particle of a pure substance (element or compound) which can exist independently. It may contain one or more atoms e.g.

Monoatomic molecules = `He`, `Ne`, `Ar`, `Kr`, `Xe`

Diatomic molecule        = `H_{2}`, `Cl_{2}`

Macromolecule              = Haemoglobin

Q.2: What is a molecular ion? How are they generated? (Lahore 2014) (Faisalabad, Sahiwal, Multan 2016)

Answer: 

When a molecule gains or loses electrons it forms a molecular ion. These are of two types:

Cationic molecular ion = `CH_4^+`, `CO^+`, `N_2^+`

Anionic molecular ion  = `N_2^-`, `O_2^{2-}`

Molecular ions can be generated by passing high energy electron beam as ` \alpha-` particles or X-rays through a gas.

Q.3: Mg atom is twice heavier than that of carbon. Explain. (Bahawalpur Board 2016)

Answer: 

Atomic mass of Mg = 24 g (1 mole)

Atomic mass of C    = 12 (1 mole)

So, it is evident that one atom of Mg is twice heavier than an atom of carbon.

Q.4: One Mg of `K_2CrO_4` has thrice the number of ions than the number of formula units when ionized in water. Explain. (Lahore 2012)

Answer: 

`K_2CrO_4\Leftrightarrow 2K^+ +CrO_4^{-2}`

One formula unit of `K_2CrO_4` ionizes into three ions i.e. two `K^+` and one `CrO_4^-`. 

So the same values are obtained taking one mg of  `K_2CrO_4`.

Q.5: 4.9 g of `H_{2}SO_{4}` when completely ionized in water have an equal number of positive and negative charges but the number of positively charged ions are twice the number of negatively charged ions. (Lahore 2012)

Answer: 

No. of moles of `H_{2}SO_{4}` = `\frac{4.9}{98}` = 0.05 moles

`H_{2}SO_4 \Leftrightarrow 2H^{+} + SO^{-2}_4`

(4.9 g)

0.05 moles            `2\times0.05`              0.05 moles

                             = 0.1 moles

4.9 g i.e 0.05 moles of `H_2SO_4` on complete ionization generates 0.1 mole of `H^+` ions and 0.05 moles of `SO_4^-` ions. So the number of positively charged ions is double the number of negatively charged ions. However 0.05 mole of `H_2SO_4` on complete ionization produce equal (0.1 moles each) of positive and negative charges.

Q.6: Calculate the number of water molecules in 9g of ice.

Answer: 

Given mass of water molecules (m)  = 9 g

Molar mass of water molecules (M)  = 18 g

        Avogadro’s No. (`N_A`) = `6.02\times 10^{23}`

  No. of water molecules (N) = ?

                                           N = `\frac{m}{M}N_A` i.e.

    No. of water molecules = `\frac{GivenMass}{MolarMass}\times N_A`

                                           = `\frac{9}{18}\times6.02\times 10^{23}`

                                           = `0.50\times6.02\times 10^{23}`

                                           = `3.01\times 10^{23}` molecules

Q.7: Calculate mass in grams of 2.74 moles of `KMnO_4`.

Answer: 

No. of moles of `KMnO_4` = n = 2.74 moles

            Mass of `KMnO_4` = m = ?

One mole of `KMnO_4`(M) = `39+55+(16\times4)=158 g`

       Mass of `KMnO_4` (m) = `n\times M`

                                           = `2.74\times 158`

                                           = 432.92 g

Q.8: Calculate mass in grams of 5.136 moles of `Ag_2CO_3`.

Answer: 

Molecular mass of `Ag_2CO_3` = M = `(108\times2)+12+48`

                                        = 276

     No. of moles of `Ag_2CO_3` = n = 5.136 moles

                                        = m = `n\times m`

                                        = `5.136\times276=14175.5g`

Q.9: How many moles are present in 52 g of Aspartame `(C_{14}H_{18}N_2O_5)`?

Answer: 

Molecular mass of Aspartame `C_{14}H_{18}N_2O_5` = 

         M = `(12\times14)+(1\times18)+(14\times2)+(16\times5)`

                       = 294 g

   Given mass = m = 52 g

       No. of moles = ?

                          n = `\frac{m}{M}`

                             = `\frac{52}{294}`

                             = 0.177 mole

Q.10: What is the mass in grams of 10.122 moles of Aspartame `(C_{14}H_{18}N_2O_5)`?

Answer: 

    Molar mass of Aspartame = M 

                = `C_{14}H_{18}N_2O_5`

                =`(12\times14)+(1\times18)+(14\times2)+(16\times5)`

                = 294 g

No. of moles of Aspartame  = n = 10.122 moles

                       Given mass (m) = ?

                                            M = `n\times m`

                                                 = `10.122\times294`

                                                 = 2975.8 g

Note: Remaining Short Questions and Answers of this chapter (FSc. Part-I Chemistry Chapter No.1: Basic Concepts) will be uploaded soon.

Attempt MCQs | Quiz of FSc Part-I, Chemistry Chapter No.1: Basic Concepts