
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
581.4k+ 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
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 English: Engaging Questions & Answers for Success

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

The largest wind power cluster is located in the state class 11 social science CBSE

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Which among the following are examples of coming together class 11 social science CBSE

Can anyone list 10 advantages and disadvantages of friction

