Tag: some basic concepts of chemistry

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Questions Related to some basic concepts of chemistry

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

 Rhenium (Re) consists of $37.1$% $185$ Re and $62.9$% $187$ Re. Calculate the relative atomic mass?

  1. $185.6$
  2. $185.9$
  3. $186.3$
  4. $186.1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Relative atomic mass is calculated as (abundance1 * mass1 + abundance2 * mass2) / 100. For Rhenium: (37.1 * 185 + 62.9 * 187) / 100 = (6863.5 + 11762.3) / 100 = 18625.8 / 100 = 186.258, which rounds to 186.3.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

The relative atomic mass of an atom is:

  1. measured in atomic mass units (u)

  2. based on the mass of 1 atom of carbon-12

  3. different for different isotopes of an element

  4. all of the above are true
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Relative atomic mass is the mass of an atom measured relative to 1/12th the mass of 1 atom of C-12 isotope which is also known as atomic mass unit or amu(u).

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

Boron found in nature has an atomic weight of 10.811 and is made up of the isotopes $\displaystyle { B }^{ 10 }$ (mass 10.013 amu) and $\displaystyle { B }^{ 11 }$ (mass 11.0093). What percentage of naturally occurring boron is made up of $\displaystyle { B }^{ 10 }$ and $\displaystyle { B }^{ 11 }$, respectively?

  1. 30 : 70

  2. 25 : 75

  3. 20 : 80

  4. 15 : 85
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the abundance of $B^{11}$ be $x$% and $B^{10}$ be $(100-x)$%

Average atomic mass= [Atomic mass of $B^{11} \times$ abundance + Atomic mass of $B^{10}\times$ abundance]$/100$ 
$\Rightarrow 10.811=\cfrac { 11.0093\times x(percent)+10.013\times (100-x)(percent) }{ 100 } $
$\Rightarrow 10.811\times 100= 11.0093x$%$+1001.3-10.013x$%
$\Rightarrow 1081.1=0.9963x$%$+1001.3$
$\Rightarrow 1081.1-1001.3=0.9963x$%
$\Rightarrow 79.8=0.9963x$%
$\Rightarrow x$%=$\cfrac {79.8}{0.9963}$
$\therefore x$%=$80$
$(100-x)$%=$20$
$\therefore$ Natural abundance of $B^{10}=20$
    Natural abundance of $B^{11}=80$
Ratio= $20:80$

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

One part of an element (A) combines with two parts of another element (B). Six parts of element (C) combines with 4 parts of (B). If (A) and (C) combine together, then the ratio of their weights will be governed by:

  1. law of definite proportion

  2. law of multiple proportion

  3. law of reciprocal proportion

  4. law of conservation of mass
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Law of reciprocal proportion states that when two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so will be the same or some simple multiple of the mass ratio in which they combine with each other.

Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Different proportions of oxygen in the various oxides of nitrogen, prove the law of:

  1. reciprocal proportions

  2. multiple proportions

  3. constant proportions

  4. conservation of mass
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Different proportion of oxygen in the various oxides of nitrogen proves the law of multiple proportions, which states:

when two elements combine in more than one proportion to form one or more compounds, the weight of one element that combine with the given weight of other elements are in the ratio of small whole number.


Multiple choice law of reciprocal proportion laws of chemical combination basic concepts of chemistry some basic concepts of chemistry chemistry

Which one of the following sets of compound correctly illustrates the law of reciprocal proportions?

  1. $P _2O _3, PH _3, H _2O$
  2. $P _2O _5, PH _3, H _2O$
  3. $N _2O _5, NH _3, H _2O$
  4. $N _2O, NH _3, H _2O$
  5. $NO _2, NH _3, H _2O$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In $PH _3$, the ratio by weight of $P:H=31:3$

For $H _2O, O:H= 16:2=8:1$
Keeping the weight of $H(=1)$ fixed, $P:O=\cfrac {31}{3}:\cfrac {8}{1}=31:24\longrightarrow (1)$
In $P _2O _5, P:O$ is $(2 \times 31):(5 \times 16)$
$=62:80$ or $31:40 \longrightarrow (2)$
Keeping the weight of $P(=31)$ fixed in equation (1) & (2), the ratio of oxygen is $24:40$ or $3:5$ which is a simple ratio.