Tag: introduction to analytical chemistry

Questions Related to introduction to analytical chemistry

0.05 mole of $LiAlH _4$ in ether solution was placed in a flask containing 74g (1 mole ) of t-butyl alcohol.The product $LiAlHC _{12}H _{27}O _3$ weighed 12.7 g. If Li atoms are conserved, the percentage yield is : (Li=7, Al=27, H=1, C=12, O=16).

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

  1. 25%

  2. 75%

  3. 100 %

  4. 15 %

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Correct Option: C

$1.25$ g of sample of limestone on heating gives $0.44$ g carbon dioxide. The percentage purity of $CaCO _3$ in limestone is:

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

  1. $75\%$

  2. $85\%$

  3. $90\%$

  4. $80\%$

Show answer
Correct Option: D
Explanation:

$CaCO _3$ $\quad \underrightarrow \Delta\quad  CaO +CO _2$

Moles of $CO _2$ produced =$\cfrac{0.44}{44}$$=0.01$ moles
Moles of pure $CaCO$$ _3$ required $=0.01$ moles
$=0.01 \times 100g$
$=1g$
Therefore $\%$ purity of $CaCO$$ _3$  in limestone =$\cfrac{1}{1.25}\times 100$
$=80\%$

$50\ g$ of an impure calcium carbonate sample decomposes on heating to give carbon dioxide and $22.4\ g$ calcium oxide. The percentage purity of calcium carbonate in the sample is:

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

  1. $60\%$

  2. $80\%$

  3. $90\%$

  4. $70\%$

Show answer
Correct Option: B
Explanation:

$CaCO _3\overset { \Delta  }{ \rightleftharpoons  } CaO+CO _2$

Let pure sample of $CaCO _3=x$ $grams$
Mass of $CaO$ produced after decomposition $=22.4$ $g$
Molar mass of $CaCO _3=100$ ${g/mol}$
and, Molar mass of $CaO=56$ $g$
If $100\%$ is pure, then
$100$ $g$ $CaCO _3\longrightarrow 56$ $g$ of $CaO$
Also,$y$ $g$ of $CaCO _3\longrightarrow 22.4$ $g$ of $CaO$
Dividing these two,
$\cfrac{100}{y}=\cfrac{56}{22.4}$
$y=40$ $grams$
$\therefore$ Percentage of purity $=\cfrac{Mass\quad of \quad pure\quad sample}{Total\quad mass\quad of\quad impure}\times 100=\cfrac{40}{50}\times 100=80\%$

Statement (A):The constant temperature at which a solid change into liquid state by absorbing heat energy is called melting point.
Statement (B): The constant temperature at which a liquid changes into gaseous state by absorbing heat energy is called point.
Statement (C): The melting and boiling points of a pure substance is always a constant and is a good measure of finding the purity of the substance.
chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

  1. All the statement are correct

  2. All the statement are incorrect

  3. A, B are correct and C is incorrect

  4. A, B are incorrect and C is correct

Show answer
Correct Option: A
Explanation:

$A\rightarrow $ Its defination of melting point.

$B\rightarrow $ Its defination of boiling point. 
$C\rightarrow $ Pure substance have constant boiling and melting point. 
$\therefore $ All are true.

If the percentage yield of the $1 st$ step is $80\% $ and that of the $2nd $ step is $75\% $, then what is the expected overall percentage yield for producing $CaO _3$ from $CaCl _{2} $?

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

  1. $50\%$

  2. $70\%$

  3. $55\%$

  4. $60\%$

Show answer
Correct Option: D
Explanation:
${ CaCl } _{ 2 }\xrightarrow [  ]{ { 1 }^{ st }step } \times \xrightarrow [  ]{ { 2 }^{ nd }step } { CaCO } _{ 3 }$
Let $100$ unit of $CaCl _2$ taken
amount of $\times $ produced $=80\ unit$
amount of $CaCO _3$ produced $=\dfrac {80\times 75}{100}=60$
Hence $\%$ yield of $CaCO _3$ by $CaCl _2=60\%$

What is the purity of concentrated $H _2SO _4$ solution $(d=1.8gm/mol)$ if $5\text{ ml}$ of these solution is neutralized by $84.5 \text{ ml}$ of $2N \text{ NaOH}$ solution. 

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

  1. $93 \text {%}$

  2. $94.6 \text {%}$

  3. $92.12 \text {%}$

  4. $91.5 \text {%}$

Show answer
Correct Option: C
Explanation:

$H _2SO _4+2NaOH \rightarrow Na _2SO _4+2H _2O$
Moles of $NaOH$ required $=\cfrac {2 \times 84.5}{1000}=0.169$
For $2 \ moles \ NaOH \rightarrow 1 \ mole \ H _2SO _4$ is used
For $\ 0.169 \ mole \ NaOH \rightarrow 0.0845 \ moles$ are used.
Mass of $H _2SO _4 \rightarrow 0.0845 \times 98=8.281 \ gm$ (Theoretical)
Mass of $H _2SO _4$ used in original reaction $\Rightarrow 5 \times 1.8 = 9 \ gm$
$\therefore$ % purity $=\cfrac {8.281}9 \times 100 = 92.12$ %

For the reaction, $2Fe(NO _3) _3+3Na _2CO _3\rightarrow Fe _2(CO _3) _3+6NaNO _3$ initially 2.5 mole of $Fe(NO _3) _2$ and 3.6 mole of $Na _2CO _3$ are taken. If 6.3 mole of $NaNO _3$ is obtained then % yield of given reaction is:

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

  1. 50

  2. 84

  3. 87.5

  4. 100

Show answer
Correct Option: C
Explanation:

$\quad \quad \quad \quad 2Fe{ (N{ O } _{ 3 }) } _{ 3 }+3{ Na } _{ 2 }{ CO } _{ 3 }\longrightarrow { Fe } _{ 2 }{ (C{ O } _{ 3 }) } _{ 3 }+6{ Na }{ NO } _{ 3 }\ initial\quad \quad 2.5\quad \quad \quad \quad \quad 3.6\quad \quad \quad \quad \quad  \quad - \quad \quad \quad \quad -\ after\quad 2.5-2.4=0.1\quad \quad 0\quad \quad \quad \quad \quad \quad 1.2 \quad \quad \quad \quad 6.3\ reaction$

As 2 moles of $Fe(NO _{3}) _{3}$ reacts with 3 moles of $Na _{2}CO _{3}$. 
Thus 2.5 moles of $Fe(NO _{3}) _{3}$ reacts with=$\cfrac{3}{2} \times 2.5$ moles of $Na _{2}CO _{3}$=3.75 moles of $Na _{2}CO _{3}$
As $Na _{2}CO _{3}$ present is 3.6 moles only. Thus, $Na _{2}CO _{3}$ is limiting reagent.
Now, 3 moles of $Na _{2}CO _{3}$ gives=6 moles of $NaNO _{3}$ 
3.6 moles of $Na _{2}CO _{3}$ gives=$\cfrac{6}{3} \times 3.6$ moles of $NaNO _{3}$
3.6 moles of $Na _{2}CO _{3}$ gives=7.2 moles of $NaNO _{3}$
But $NaNO _{3}$ obtained is 6.3 moles.
Thus % yield=$\cfrac{6.3}{7.2} \times 100=87.5 \%$

$12.5$ g of an impure sample of limestone on heating gives $4.4$ g of carbon dioxide. The percentage purity of $CaCO _{3}$ in the sample is:

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

  1. $72$%

  2. $75$%

  3. $80$%

  4. $85$%

Show answer
Correct Option: C
Explanation:

${ CaCO } _{ 3 }\overset { \Delta  }{ = } CaO+{ CO } _{ 2 }\uparrow $

$100gm$                      $44gm$
Therefore $1$ mole i.e. $100gm$ of ${ CaCO } _{ 3 }$ (lime stone) give $1$ mole i.e. $44gm$ of ${ CO } _{ 2 }$.
$4.4gm$ of ${ CO } _{ 2 }$ are produced from $\dfrac { 100\times 4.4 }{ 44 } gm$ i.e. $10gm$ of ${ CaCO } _{ 3 }$.
$\therefore$   The percentage of purity $=\left( 1-\dfrac { 12.5-10 }{ 12.5 }  \right) \times 100$% $=80$%
$\therefore$   Correct answer is $C$ $(80$%$)$.

For the complete reduction of $5.8g$ of acetone to isopropyl alcohol, the quantity of $LiAIH _{4}$ required (assuming chemical yield to be $100\%$ ) is approximately [mass:$Li=6.9,Al=27$]

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

  1. $5.8\ g$

  2. $3.8\ g$

  3. $1.9\ g$

  4. $15.2\ g$

Show answer
Correct Option: B

0.2828 g of iron wire was dissolved in excess of dilute $H _2SO _4$ and the solution was made upto 100 ml. 20 ml of this solution required 30 ml. of $\dfrac{N}{30} K _2Cr _2O _7$ solution for oxidation. Calculate % purity of iron in the wire:

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

  1. 99

  2. 95

  3. 90

  4. 85

Show answer
Correct Option: B