On heating a liter of a \[\dfrac {N}{{{\text{ }}2}}\]\[HCl\] solution, \[2.750\] gram of \[HCl\] is lost and the volume of solution becomes\[750{\text{ }}ml\]. The normality of resulting solution will be
A) $0.57$
B) \[0.075\]
C) \[0.057\]
D) \[5.7\]
Answer
590.1k+ views
Hint: In the given question the amount of volume of lost \[HCl\] in the solution has given, according to it we have to find the normality of the resulting solution. As the normality is the measure of concentration equal to the gram equivalent weight per liter of solution.
Complete answer:
The Normality is often denoted by the letter \[N\] and Some of the other units of normality are also expressed as \[eq/{L^{ - 1}}\]. The latter is often used in medical diagnosis. The common equation of normality \[N\] is equal to the gram equivalent weight which is divided by liters of solution.
According to the question for calculating the normality firstly we have to calculate the moles of hydrochloric acid.
So, Moles of \[HCl{\text{ }} = {\text{ }}1 \times \dfrac {1}{2} = 0.5\]
Therefore, the Moles of \[HCl\] lost in solution \[ = {\text{ }}\dfrac {{2.75{\text{ }}}}{{36.5}}{\text{ }} = {\text{ }}0.075\] moles
Thus, the Moles of \[HCl\] left in solution \[ = {\text{ }}0.5{\text{ }}-{\text{ }}0.075{\text{ }} = {\text{ }}0.425\]
Now, the normality of the final solution will be
\[ = {\text{ }}\dfrac {{number{\text{ }}of\;moles{\text{ }}}}{{volume{\text{ }}in{\text{ }}litre}}\]
Putting the values in formula
$ = \dfrac {{0.425\;}}{{\dfrac {{750}}{{1000}}}}{\text{ }}$
So, the normality will be \[ = {\text{ }}0.57{\text{ }}N\]
Hence, the correct answer is option ‘A’.
Note: Normality in terms of chemistry is an expression that is used to measure the concentration of a solution and it is abbreviated as ‘N’ and is sometimes referred to as the equivalent concentration of a solution. Normality is also used in the reactions such as precipitation reactions to measure the number of ions which are probable to precipitate in a specific reaction.
Complete answer:
The Normality is often denoted by the letter \[N\] and Some of the other units of normality are also expressed as \[eq/{L^{ - 1}}\]. The latter is often used in medical diagnosis. The common equation of normality \[N\] is equal to the gram equivalent weight which is divided by liters of solution.
According to the question for calculating the normality firstly we have to calculate the moles of hydrochloric acid.
So, Moles of \[HCl{\text{ }} = {\text{ }}1 \times \dfrac {1}{2} = 0.5\]
Therefore, the Moles of \[HCl\] lost in solution \[ = {\text{ }}\dfrac {{2.75{\text{ }}}}{{36.5}}{\text{ }} = {\text{ }}0.075\] moles
Thus, the Moles of \[HCl\] left in solution \[ = {\text{ }}0.5{\text{ }}-{\text{ }}0.075{\text{ }} = {\text{ }}0.425\]
Now, the normality of the final solution will be
\[ = {\text{ }}\dfrac {{number{\text{ }}of\;moles{\text{ }}}}{{volume{\text{ }}in{\text{ }}litre}}\]
Putting the values in formula
$ = \dfrac {{0.425\;}}{{\dfrac {{750}}{{1000}}}}{\text{ }}$
So, the normality will be \[ = {\text{ }}0.57{\text{ }}N\]
Hence, the correct answer is option ‘A’.
Note: Normality in terms of chemistry is an expression that is used to measure the concentration of a solution and it is abbreviated as ‘N’ and is sometimes referred to as the equivalent concentration of a solution. Normality is also used in the reactions such as precipitation reactions to measure the number of ions which are probable to precipitate in a specific reaction.
Recently Updated Pages
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

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

