Tag: introduction to analytical chemistry

Questions Related to introduction to analytical chemistry

$0.5\ g$ of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in $150\ mL$ of $N/5\ H _{2}SO _{4}$ solution. Excess sulphuric acid required $20\ mL$ of $1\ N\ NaOH$ for complete neutralization. The percentage of $NH _{3}$ in the ammonium chloride is:

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $68$%

  2. $34$%

  3. $48$%

  4. $17$%

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Correct Option: B
Explanation:

$(NH _3)Cl+NaOH\rightarrow +NH _3\uparrow +H _2O$

excess $H _2SO _4$ reg. $20ml$ of $1N\,\,NaOH$
$\Rightarrow $ moles of $H _2SO _4=(20\times 1)milimoles$
                                 $=20m\,mol$
added amount of $H _2SO _4=\dfrac{N}{5}\times 150ml=30mmol$
amount of $h _2so _4$ reacted with $nh _3=0.01MOLE$
$\Rightarrow 0.01\, mole$ of $NH _3$ present in $(NH _4)Cl$
So, $\%purity=\dfrac{(0.01)\times 17}{0.5}\times 100=34\%$

A sample of $CaCO _3$ is 50% pure. On heating $1.12 L$ of $CO _2$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g

Show answer
Correct Option: A
Explanation:

Solution:- (A) $7.8 \; g$

Volume of $C{O} _{2}$ formed $= 1.12 \; L$
At STP, volume of $1$ mole of gas $= 22.4 \; L$
$\therefore$ No. of moles of $C{O} _{2}$ formed $= \cfrac{1.12}{22.4} = 0.05 \text{ mol}$
Heating of $CaC{O} _{3}$-

$CaC{O} _{3} \longrightarrow CaO + C{O} _{2}$
From the above reaction,
$1$ mole of $C{O} _{2}$ is formed on heating $1$ mole of $CaC{O} _{3}$.
Therefore,
$0.05$ mole of $C{O} _{2}$ is formed on heating $0.05$ mole of $CaC{O} _{3}$.

Molecular weight of $CaC{O} _{3} = 100 \; g$
$\therefore$ Weight of $CaC{O} _{3} = 0.05 \times 100 = 5 \; g$
As the sample was $50 \%$ pure.
Thus the $50 \%$ of the sample was heated in the form of $CaC{O} _{3}$.
Amount of sample left unreacted $= 5 \; g$
Also,
No. of moles of $CaO$ formed $= 0.05$
Molecular weight of $CaO = 56 \; g$
Weight of $CaO = 56 \times 0.05 = 2.8 \; g$
Therefore,
Amount of residue left $= 5 + 2.8 = 7.8 \; g$

In ayurvedic preparation of swarnabhasma, what purity of golden will be used?

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. $42$ % gold + silver

  2. $91$ % amalgam

  3. $24$ carat

  4. $58.5$% copper

Show answer
Correct Option: A
Explanation:

Swarna Bhasma is prepared from Gold. It is used in Ayurvedic treatment of infertility, asthma, tissue wasting, poisoning etc. This medicine should only be taken strictly under medical supervision.

In the given question option A is the correct answer.

One mole of photons is known as one Einstein of radiation. According to Stark-Einstein law of photochemical equivalence, one mole of reactant absorbs one Einstien of energy. For a photochemical reaction, a term called quantum yield is defined as:
Quantum yield $ (\phi) = \dfrac {No. \,of \,moles \,of \,reactant \,converted} {No. \,of \,Einstein \,absorbed} $
The correct statement(s) is/are:

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. for a chain reaction $\phi _{gas} >> \phi _{solution}$

  2. in a photochemical chain reaction $\phi >> 1$

  3. in a photochemical chain reaction $\phi << 1$

  4. for a chain reaction $\phi _{gas} << \phi _{solution}$

Show answer
Correct Option: A,B
Explanation:
$Quantum \ Yield= \cfrac {Number \ of \ moles \ of \ reactant \ converted}{Number \ of \ einstein \ absorbed}$
An einstein of radiation $=$ one mole of photons
For a chain reaction, $\phi _{gas} >> \phi _{solution}$
and ln photochemical chain reaction $\phi >>1$

A sample of $CaC{O _3}$ is $50\% $ pure. On heating $1.12{\text{ }}L$ of $C{O _2}$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is:

chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g

Show answer
Correct Option: A
Explanation:

No. of moles of $CO _2$ evolved $=\cfrac{1.12}{22.4}=0.05$ $moles$

$CaCO _3(s)\overset { \Delta }{ \longrightarrow } CaO\downarrow+CO _2\uparrow$
                             $0.05$        $0.05$ $moles$
So, $0.05$ $moles$ of $CaCO _3$ have  reacted.
Mass $=0.05\times 100=5$ $gm=50\%$ of $CaCO _3$ sample
Total weight $=2\times 5$ $gm=10$ $gm$
Residue left by $CaCO _3=5$ $gm$
Residue left by $CaO=56\times 0.05=2.8$ $gm$
Toatl residue $=5+2.8=7.8$ $gm$

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 $\text{C}$