Tag: chemistry

Questions Related to chemistry

Potash alum contains ___________ number of water molecules.

chemistry substances in common use percent composition water of crystallisation percentage composition and empirical formula

  1. 24

  2. 22

  3. 20

  4. 25

Show answer
Correct Option: A
Explanation:

Potash Alum $(K _2SO _4 Al _2 (SO _4) _3 .24H _2O)$ is a colourless, odourless and crystalline substance. Its crystals are prepared by mixing concentrated solutions of potassium sulphate and aluminium sulphate. It is soluble in water. It contains 24 molecules of water.

Copper sulphate pentahydrate is commonly called as____, whereas ferrous sulphate heptahydrate is commonly called as _____ .

chemistry substances in common use percent composition water of crystallisation percentage composition and empirical formula

  1. white vitriol, blue vitriol

  2. green vitriol, blue vitriol

  3. blue vitriol, green vitriol

  4. blue vitriol, white vitriol

Show answer
Correct Option: C
Explanation:

Copper sulphate pentahydrate ($CuSO _4.5H _2O$) commonly known as blue vitrol contains 5 water molecules which gives it a blue color. Similarly, the green color of ferrous sulphate heptahydrate ($FeSO _4.7H _2O$) is due to 7 molecules water present in it. This water is responsible for the geometric shape and color of the crystals.

Compound Q contains $40\%$ carbon by mass.
What could Q be?
$1$. glucose, $C _6H _{12}O _6$
$2$. starch, $(C _6H _{10}O _5) _n$
$3$. sucrose, $C _{12}H _{22}O _{11}$.

chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

  1. $1, 2$ and $3$ are correct

  2. $1$ and $2$ only are correct

  3. $2$ and $3$ only are correct

  4. $1$ only is correct

Show answer
Correct Option: D
Explanation:
$\text{percentage mass of C}=\dfrac{\text{mass of C X 100%}}{\text{total mass}}$

$\text{1. percentage of C in glucose=40%}$
$\text{2. percentage of C in starch=44%}$
$\text{3. percentage of C in sucrose=42%}$

What is % composition of a substance?

chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

  1. Sum of all the components.

  2. % composition of the sum of two components.

  3. % of the total mass of a substance.

  4. None of the above.

Show answer
Correct Option: C
Explanation:

The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component.

What is the percent of composition of the compound that forms when $222.7g$ of N combines compleletly with $77.4g$ of O?

chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

  1. $70%, 30%$

  2. $80%, 20%$

  3. $74.2%, 25.8%$

  4. $72.2%, 27.8%$

Show answer
Correct Option: C
Explanation:

The mass composition$:$
Total mass$: 222.7+77.4=300.1$
Mass percentage of N$:$
$222.7300.1=74.25$
Mass percentage of O$:$
$100-74.2=25.8%$

Calculate the percent composition of carbon in $C _6H _{12}O _6$ :

chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

  1. $50%$

  2. $40%$

  3. $35.3%$

  4. $22.1%$

Show answer
Correct Option: B
Explanation:

Molar mass of compound $: 6(12.0)+12(1.01)+6(16.0)=180g/mol$
Mass due to carbon $=6(12.0)=72.06g/mol$
$%$ composition of $C=72/180\times 100=40%$

The formula for % composition of a compound is:

chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

  1. molar mass of a compound/mass due to specific component $\times$ 100

  2. mass due to specific component $\times$ 100

  3. mass due to specific component/molar mass of a compound $\times$ 100

  4. molar mass of a specific component/temperature $\times$ 100

Show answer
Correct Option: C
Explanation:

To calculate the per cent composition of a component in a compound: Find the molar mass of the compound by adding up the masses of each atom in the compound using the periodic table or a molecular mass calculator. Calculate the mass due to the component in the compound you are for which you are solving by adding up the mass of these atoms. Divide the mass due to the component by the total molar mass of the compound and multiply by 100.

% $\text{composition of a compound}=\cfrac{\text{ Mass due to specific component}}{\text{molar mass of a compound}}\times 100$

Hence, the correct option is $\text{C}$

% composition requires ................ of the compound :

chemistry introduction to analytical chemistry percent composition water of crystallisation percentage composition and empirical formula

  1. molar mass

  2. temperature

  3. atmospheric pressure

  4. both a and b

Show answer
Correct Option: A
Explanation:

The present composition (by mass) of a compound can be calculated by dividing the mass of each element by the total mass of the compound. i.e. by calculating the $molar$ $mass$ of the compound.

Lanthanides and actinides resemble in:

the actinoids actinoids comparison of lanthanoids and actinoids d- and f-block elements chemistry

  1. electronic configuration

  2. oxidation state

  3. ionization energy

  4. formation of complexes

Show answer
Correct Option: A
Explanation:

Lanthanides and actinides resemble in electronic configuration:

$(n-2)f^{1-14}(n-1)d^{1-10}ns^{1-2}$

More number of oxidation states are exhibited by the actinoids than by lanthanoids. The main reason for this is:

chemistry d- and f-block elements comparison of lanthanoids and actinoids actinoids the actinoids

  1. greater metallic character of the lanthanoids than that of the corresponding actinoids.

  2. more active nature of the actinoids.

  3. more energy difference $5f$ and $6d$-orbitals than that between $4f$ and $5d$-orbitals.

  4. lesser energy difference between $5f$ and $6d$-orbitals than that between $4f$ and $5d$-orbitals.

Show answer
Correct Option: D
Explanation:

Actinoids display more oxidation states because of very small energy gap between 5f, 6d and 7s sub shells. Thus, the outermost electrons get easily excited to the  higher energy level giving variable oxidation state.

Hence,option D is correct.