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Atoms and Molecules Test - 53

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Atoms and Molecules Test - 53
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  • Question 1
    1 / -0
    Calculate the number of atoms of sulphur present in $$0.5$$ moles of $${ Na }_{ 2 }{ S }_{ 2 }{ O }_{ 3 }$$:
    Solution
    $$1$$ mole of $$Na_2S_2O_3$$ contains Avogadro's number of molecules which is $$ \displaystyle 6.023 \times 10^{23}$$ molecules.

    $$0.5$$ mole of $$Na_2S_2O_3$$ will contain $$ \displaystyle 0.5 \times 6.023 \times 10^{23}=3.0115 \times 10^{23}$$ molecules.

    One molecule of $$Na_2S_2O_3$$ contains $$2\ S$$ atoms.
     
    $$ \displaystyle  3.0115 \times 10^{23}$$ molecules of $$Na_2S_2O_3$$ will contain  $$ \displaystyle 2 \times 3.0115 \times 10^{23}=6.023 \times 10^{23}$$  $$S$$ atoms.
    Hence, option $$C$$ is correct.
  • Question 2
    1 / -0
    What weight of calcium contains the same number of atoms as those in $$3 g$$ of carbon?
    Solution
    Atomic weights of $$C$$ and $$Ca$$ are 12 g/mol and 40 g/mol respectively.
    3 g $$C  \displaystyle = \dfrac {3 \ g}{12 \ g/mol}= 0.25 \ mol C$$.
    Number of moles of $$Ca  \displaystyle =$$ number of moles of $$C  \displaystyle = $$ 0.25.
    Weight of $$Ca  \displaystyle = 3 \ mol \times 40 \ g/mol = 10 \ g$$.
    Note: Since $$C$$ and $$Ca$$ samples have same number of atoms, they will have same number of moles.
  • Question 3
    1 / -0
    What is the approximate molecular mass of dry air containing 78% $$N_2$$ and 22% $$O_2$$?
    Solution
    The molecular mass of dry air is given by-

    $$ \displaystyle = \dfrac {  [ \text { Molecular mass of }N_2   \times \text { Percentage of }N_2  ]   + [ \text { Molecular mass of }O_2   \times \text { Percentage of }O_2  ]   }  {    \text { Percentage of }N_2   +    \text { Percentage of }O_2  }   $$

     $$ \displaystyle = \dfrac {  [ \text { 28 }   \times \text { 78 }  ]   + [ \text { 32 }   \times \text { 22 }  ]   }  {    \text { 78   } +    \text {22  }  }   $$

    $$ \displaystyle = \text {28.88 g/mol}$$

    Hence, option $$A$$ is correct.
  • Question 4
    1 / -0
    Identify the pair of substances having the same formula unit mass.
    Solution
    Carbon dioxide and nitrous oxide have the same formula unit mass.
    Formula unit mass for carbon dioxide $$ \displaystyle CO_2 = 12 +2(16)=12+32=44\ u$$
    Formula unit mass for nitrous oxide $$ \displaystyle N_2O = 2(14)+16=28+16=44\ u$$

    Let's look at the formula unit mass of the other given compounds:
    Formula unit mass for calcium chloride $$ \displaystyle CaCl_2 = 40+2(35.5)=40+71=111\ u$$
    Formula unit mass for potassium carbonate $$ \displaystyle K_2CO_3 = 2(39)+12+3(16)=78+12+48=138\ u$$
    Formula unit mass for calcium oxide $$ \displaystyle CaO = 40+16=56\ u$$
    Formula unit mass for hydrochloric acid $$ \displaystyle HCl = 1+35.5=36.5\ u$$
    Formula unit mass for carbon monoxide $$ \displaystyle CO = 12+16=28\ u $$
    Formula unit mass for ammonia $$ \displaystyle NH_3 = 14+3(1)=17\ u$$
  • Question 5
    1 / -0
    The heaviest among the following is:
    Solution
    (A) $$50$$ g of $$\displaystyle Fe$$

    (B) 5 moles of $$\displaystyle N_2  = 5 \: mol \times 28 \: g/mol = 140 $$ g.

    (C) 0.1 g atom of $$Ag$$ $$\displaystyle = 0.1 \: g \: atom \times 108 \: g/ \:  atom = 10.8 $$ g 

    (D) $$0.1$$ g $$\displaystyle Ca$$

    Hence, 5 moles of $$\displaystyle N_2  $$ are heaviest.
  • Question 6
    1 / -0
    What is the mass of one molecule of oxygen?
    Solution

    Explanation:

    The molecular mass of oxygen ($$O_2$$) is $$32\ grams$$.

    $$\therefore$$ $$6.023 \times 10^{23}$$ molecules of oxygen ($$O_2$$) have mass $$=32\ grams$$.

    $$\therefore$$ One molecule of oxygen ($$O_2$$) will have mass $$=\dfrac{32}{6.023\times 10^{23}}=5.31\times 10^{-23}\ grams$$.

    Final Answer: One molecule of oxygen ($$O_2$$) have mass $$=5.31\times 10^{-23}\ grams$$. Hence option $$A$$ is correct.

  • Question 7
    1 / -0
    Calculate the number of atoms present in $$71 g$$ of $${ Cl }_{ 2 }$$.
    Solution
    Weight of   $$ \displaystyle Cl_2 = 71$$ g

    Molecular weight of $$ \displaystyle Cl_2 $$ $$ \displaystyle =71$$ g/mol

    Number of moles of $$ \displaystyle Cl_2 $$ $$ \displaystyle = \dfrac {71 \ g}{71 \ g/g atom}=1$$ moles.

    The Avogadro's number is $$ \displaystyle 6.023 \times 10^{23}$$ molecules/mol.

    The number of atoms of $$ \displaystyle Cl_2 $$ $$ \displaystyle =2 $$  atoms/molecule $$ \times 1$$ mol  $$ \displaystyle  \times 6.023 \times 10^{23}$$ molecules/mol
    $$ \displaystyle =1.205 \times 10^{24}$$ atoms
  • Question 8
    1 / -0
    Who was the first scientist to propose that all matter was made of small spheres called atoms?
    Solution
    John Dalton was the first scientist to propose that all matter was made of small spheres called atoms. The main points of Dalton's Atomic theory given by John Dalton are:-
    (a) Elements are extremely small particles called atoms.
    (b) Atoms of a given element are identical in size, mass, and other properties, 
    (c) Atoms cannot be sublimed, created, or destroyed.
    (d) Atoms of different elements combine in simple whole-number ratios to form chemical compounds.
    (e) In chemical reactions, atoms are combined, separated, or rearranged.
    Option $$A$$ is correct.
  • Question 9
    1 / -0
    A smallest particle of a compound that can have the independence existence and that can retain all the properties of that compound is called as:
    Solution
    A smallest part of the compound that can have the independence existence and that can retain all the properties of that compound is called as molecule. Ex: $$H_2,Cl_2$$ 
  • Question 10
    1 / -0
    Which of the following term is denoted as Formula Mass Unit?
    Solution
    Formula unit mass is the sum of the atomic masses of all the ions in the ionic compound. Formula unit mass is assigned to the ionic compounds.
    Hence option C is correct.
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