Answer
Verified
285k+ views
Hint: Moles is the concept that represents the number of atoms present in one mole, which is equal to $ 6.023 \times {10^{23}} $ known as Avogadro’s number. As one mole consists of Avogadro's number of atoms, by multiplying the given number of moles with Avogadro’s number gives the number of atoms.
Complete answer:
Given reaction is $ HN{O_3} + NaOH \to NaN{O_3} + {H_2}O $
One mole consists of $ 6.023 \times {10^{23}} $ number of molecules, $ 0.55 $ moles of $ HN{O_3} $ consists of $ 6.023 \times {10^{23}} \times 0.55 = 3.312 \times {10^{23}} $ molecules. As there are three oxygen atoms in one molecule of $ HN{O_3} $ .
The oxygen atoms will be $ 3 \times 3.312 \times {10^{23}} = 9.93 \times {10^{23}} $
Thus, $ 0.55 $ moles of $ HN{O_3} $ consist of $ 9.93 \times {10^{23}} $ oxygen atoms.
The number of moles is the ratio of mass to molar mass.
Given that the $ 0.365g $ of $ NaOH $ is reacted, the molar mass of $ NaOH $ is $ 40gmo{l^{ - 1}} $ .
The number of moles of $ NaOH $ will be $ \dfrac{{0.365}}{{40}} = 0.091moles $
As one mole of $ NaOH $ reacts with one mole of $ HN{O_3} $ to produce one mole of $ NaN{O_3} $
Thus, $ 0.091moles $ of $ NaOH $ produces $ 0.091moles $ of $ NaN{O_3} $
The molar mass of $ NaN{O_3} $ is $ 85gmo{l^{ - 1}} $
Thus, the mass of $ NaN{O_3} $ will be $ 0.091mol \times 85gmo{l^{ - 1}} = 0.7335g $
If $ 0.365g $ of $ NaOH $ is reacted with excess of $ HN{O_3} $ the maximum mass of $ NaN{O_3} $ which could be recovered is $ 0.7335g $
Given that the initial volume is $ 12.05c{m^3} $ with initial concentration of $ 0.2065M $ aqueous $ NaOH $ solution is titrated to the end-point by a final volume of $ 25.05c{m^3} $ of $ HN{O_3} $ solution.
Substitute these values in the formula, $ {M_1}{V_1} = {M_2}{V_2} $
$ 12.05c{m^3} \times 0.2065M = {M_2} \times 25.05c{m^3} $
Further simplification, $ {M_2} = 0.099M $
Thus, $ 12.05c{m^3} $ of a $ 0.2065M $ aqueous $ NaOH $ solution is titrated to the end-point by $ 25.05c{m^3} $ of $ HN{O_3} $ solution the concentration of the $ HN{O_3} $ will be $ 0.099M $ .
Note:
While calculating the number of moles, the molar mass should be taken exactly and the balanced chemical equation only must be considered. Based on this balanced equation only, the mole ratio is taken and according to this the moles of product can be calculated.
Complete answer:
Given reaction is $ HN{O_3} + NaOH \to NaN{O_3} + {H_2}O $
One mole consists of $ 6.023 \times {10^{23}} $ number of molecules, $ 0.55 $ moles of $ HN{O_3} $ consists of $ 6.023 \times {10^{23}} \times 0.55 = 3.312 \times {10^{23}} $ molecules. As there are three oxygen atoms in one molecule of $ HN{O_3} $ .
The oxygen atoms will be $ 3 \times 3.312 \times {10^{23}} = 9.93 \times {10^{23}} $
Thus, $ 0.55 $ moles of $ HN{O_3} $ consist of $ 9.93 \times {10^{23}} $ oxygen atoms.
The number of moles is the ratio of mass to molar mass.
Given that the $ 0.365g $ of $ NaOH $ is reacted, the molar mass of $ NaOH $ is $ 40gmo{l^{ - 1}} $ .
The number of moles of $ NaOH $ will be $ \dfrac{{0.365}}{{40}} = 0.091moles $
As one mole of $ NaOH $ reacts with one mole of $ HN{O_3} $ to produce one mole of $ NaN{O_3} $
Thus, $ 0.091moles $ of $ NaOH $ produces $ 0.091moles $ of $ NaN{O_3} $
The molar mass of $ NaN{O_3} $ is $ 85gmo{l^{ - 1}} $
Thus, the mass of $ NaN{O_3} $ will be $ 0.091mol \times 85gmo{l^{ - 1}} = 0.7335g $
If $ 0.365g $ of $ NaOH $ is reacted with excess of $ HN{O_3} $ the maximum mass of $ NaN{O_3} $ which could be recovered is $ 0.7335g $
Given that the initial volume is $ 12.05c{m^3} $ with initial concentration of $ 0.2065M $ aqueous $ NaOH $ solution is titrated to the end-point by a final volume of $ 25.05c{m^3} $ of $ HN{O_3} $ solution.
Substitute these values in the formula, $ {M_1}{V_1} = {M_2}{V_2} $
$ 12.05c{m^3} \times 0.2065M = {M_2} \times 25.05c{m^3} $
Further simplification, $ {M_2} = 0.099M $
Thus, $ 12.05c{m^3} $ of a $ 0.2065M $ aqueous $ NaOH $ solution is titrated to the end-point by $ 25.05c{m^3} $ of $ HN{O_3} $ solution the concentration of the $ HN{O_3} $ will be $ 0.099M $ .
Note:
While calculating the number of moles, the molar mass should be taken exactly and the balanced chemical equation only must be considered. Based on this balanced equation only, the mole ratio is taken and according to this the moles of product can be calculated.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE