Answer
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Hint: Normality is generally used to measure the concentration of a solution. It is represented by the letter N and sometimes referred to as the equivalent concentration of a solution. Units of normality expressed as $eq{{L}^{-1}}$ or $meq{{L}^{-1}}$.
Complete Solution :
Normality can also be defined as the number of gram or mole equivalents of solute present in one litre of a solution. Where equivalent represents the number of moles of reactive units in a compound.
Normality is calculated by the formula: $N=\dfrac{Weight~of~solute}{equivalent~weight\times volume}$
Normality of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O$ solution = $\dfrac{24.8}{248\times 1}=0.1N$
Applying the normality equation:
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
$Volume~of~A{{S}_{2}}{{O}_{3}}solution~in~NaHC{{O}_{3}}\times Normality$ = $Volume~of~iodine~solution\times Normality$
- We have to find out normality of $A{{S}_{2}}{{O}_{3}} solution in NaHC{{O}_{3}}$, let us suppose it as N
$25\times N = 22.4\times 0.1$
$N=\dfrac{22.4\times 0.1}{25} = 0.0896$
Amount of $A{{s}_{2}}{{O}_{3}}$ present in 250 ml of solution is given by:
$N\times \dfrac{Equivalent~mass~of~A{{s}_{2}}{{O}_{3}}}{1000}\times 250$
$0.0896\times \dfrac{198}{4\times 1000}\times 250 = 1.1088g$
Percentage of $A{{s}_{2}}{{O}_{3}}$ = $\dfrac{1.1088}{12}\times 100=9.24$
Percentage of arsenious oxide in the sample is 9.24%.
Note: Arsenious oxide is an inorganic compound. It is generally used as a medication to treat a type of cancer called acute promyelocytic leukemia in which it is by injection directly to the vein. It is also used in the manufacture of wood preservatives, pesticides etc.
Complete Solution :
Normality can also be defined as the number of gram or mole equivalents of solute present in one litre of a solution. Where equivalent represents the number of moles of reactive units in a compound.
Normality is calculated by the formula: $N=\dfrac{Weight~of~solute}{equivalent~weight\times volume}$
Normality of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}.5{{H}_{2}}O$ solution = $\dfrac{24.8}{248\times 1}=0.1N$
Applying the normality equation:
${{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}$
$Volume~of~A{{S}_{2}}{{O}_{3}}solution~in~NaHC{{O}_{3}}\times Normality$ = $Volume~of~iodine~solution\times Normality$
- We have to find out normality of $A{{S}_{2}}{{O}_{3}} solution in NaHC{{O}_{3}}$, let us suppose it as N
$25\times N = 22.4\times 0.1$
$N=\dfrac{22.4\times 0.1}{25} = 0.0896$
Amount of $A{{s}_{2}}{{O}_{3}}$ present in 250 ml of solution is given by:
$N\times \dfrac{Equivalent~mass~of~A{{s}_{2}}{{O}_{3}}}{1000}\times 250$
$0.0896\times \dfrac{198}{4\times 1000}\times 250 = 1.1088g$
Percentage of $A{{s}_{2}}{{O}_{3}}$ = $\dfrac{1.1088}{12}\times 100=9.24$
Percentage of arsenious oxide in the sample is 9.24%.
Note: Arsenious oxide is an inorganic compound. It is generally used as a medication to treat a type of cancer called acute promyelocytic leukemia in which it is by injection directly to the vein. It is also used in the manufacture of wood preservatives, pesticides etc.
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