
$5{\text{g}}$ sample of urea when heated with \[{\text{NaOBr}}\] and \[{\text{NaOH}}\] gives \[1120{\text{mL}}\] of nitrogen at STP. The percentage purity of sample is:
A. \[50\% \]
B. \[40\% \]
C. \[60\% \]
D. \[30\% \]
Answer
573.9k+ views
Hint:Percentage purity of a substance can be measured by dividing mass of pure compound by total mass of sample multiplied by 100. Using balanced chemical reactions to determine how much reactant or product is consumed or produced in a chemical reaction. Coefficients in chemical equations represent the mole ratios.
Complete step by step answer:
Stoichiometry is the branch of chemistry that deals with the quantities of chemicals. It can be used to tell whether the elements took the form of solid, liquid or gaseous form.
The chemical reaction is given below:
\[{\text{C}}{{\text{H}}_2}{\text{CON}}{{\text{H}}_2} + 3{\text{NaOBr}} + 2{\text{NaOH}} \to {{\text{N}}_2} + 3{\text{NaBr}} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + 3{{\text{H}}_2}{\text{O}}\]
Urea sodium hypobromite
Molar volume of any ideal gas at \[273.15{\text{K}}\] temperature and \[1{\text{atm}}\] pressure is equal to \[22.4{\text{L}}\] or \[22400{\text{mL}}\].
Here, \[1{\text{mol}}\] of urea gives \[1{\text{mol}}\] of nitrogen.
i.e. \[22400{\text{mL}}\] of \[{{\text{N}}_2}\] is obtained from \[60{\text{g}}\] of urea at STP.
Similarly \[1{\text{mL}}\] of \[{{\text{N}}_2}\] is obtained from \[\dfrac{{60{\text{g}}}}{{22400{\text{mL}}}} = 2.68 \times {10^{ - 3}}{\text{gm}}{{\text{L}}^{ - 1}}\]
Therefore, \[1120{\text{mL}}\]of \[{{\text{N}}_2}\] is obtained from \[\left( {2.68 \times {{10}^{ - 3}}{\text{gm}}{{\text{L}}^{ - 1}}} \right) \times 1120{\text{mL = 3}}{\text{.0016g}} \sim {\text{3g}}\]
It is calculated as \[5{\text{g}}\] sample contains \[3{\text{g}}\] urea.
\[1{\text{g}}\] sample contains \[\dfrac{3}{5} = 0.6{\text{g}}\]
Since purity is measured in percentage, we can calculate how much urea is contained in \[100{\text{g}}\] sample
\[100{\text{g}}\] sample contains $0.6 \times 100 = 60{\text{g}}$ of urea.
Therefore purity percentage is \[60\% \]
Additional information- All reactants may not convert to products. Most of the reactions are less than 100% complete. Percentage purity and percentage yield are different. Percentage yield is a measure of what percentage of reactants have become products.
Note:
Not all chemicals in chemical reactions are pure. Stoichiometry helps to find out the composition and general proportions of how to combine things to get them react completely. We cannot directly convert these chemicals into moles in a calculation. Percentage purity is the percentage of a pure compound in an impure sample.
Complete step by step answer:
Stoichiometry is the branch of chemistry that deals with the quantities of chemicals. It can be used to tell whether the elements took the form of solid, liquid or gaseous form.
The chemical reaction is given below:
\[{\text{C}}{{\text{H}}_2}{\text{CON}}{{\text{H}}_2} + 3{\text{NaOBr}} + 2{\text{NaOH}} \to {{\text{N}}_2} + 3{\text{NaBr}} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + 3{{\text{H}}_2}{\text{O}}\]
Urea sodium hypobromite
Molar volume of any ideal gas at \[273.15{\text{K}}\] temperature and \[1{\text{atm}}\] pressure is equal to \[22.4{\text{L}}\] or \[22400{\text{mL}}\].
Here, \[1{\text{mol}}\] of urea gives \[1{\text{mol}}\] of nitrogen.
i.e. \[22400{\text{mL}}\] of \[{{\text{N}}_2}\] is obtained from \[60{\text{g}}\] of urea at STP.
Similarly \[1{\text{mL}}\] of \[{{\text{N}}_2}\] is obtained from \[\dfrac{{60{\text{g}}}}{{22400{\text{mL}}}} = 2.68 \times {10^{ - 3}}{\text{gm}}{{\text{L}}^{ - 1}}\]
Therefore, \[1120{\text{mL}}\]of \[{{\text{N}}_2}\] is obtained from \[\left( {2.68 \times {{10}^{ - 3}}{\text{gm}}{{\text{L}}^{ - 1}}} \right) \times 1120{\text{mL = 3}}{\text{.0016g}} \sim {\text{3g}}\]
It is calculated as \[5{\text{g}}\] sample contains \[3{\text{g}}\] urea.
\[1{\text{g}}\] sample contains \[\dfrac{3}{5} = 0.6{\text{g}}\]
Since purity is measured in percentage, we can calculate how much urea is contained in \[100{\text{g}}\] sample
\[100{\text{g}}\] sample contains $0.6 \times 100 = 60{\text{g}}$ of urea.
Therefore purity percentage is \[60\% \]
Additional information- All reactants may not convert to products. Most of the reactions are less than 100% complete. Percentage purity and percentage yield are different. Percentage yield is a measure of what percentage of reactants have become products.
Note:
Not all chemicals in chemical reactions are pure. Stoichiometry helps to find out the composition and general proportions of how to combine things to get them react completely. We cannot directly convert these chemicals into moles in a calculation. Percentage purity is the percentage of a pure compound in an impure sample.
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