
0.115 g of pure sodium metal was dissolved in 500 ml distilled water. The normality of the above solution, whose resulting volume is 400 ml, would be;
A. 0.010 N
B. 0.0115 N
C. 0.0125 N
D. 0.046 N
Answer
160.8k+ views
Hint: Normality of a solution is the number of gram equivalents of the solute dissolved per litre of a given solution. It is signified by N.
Formula Used:
\[{\rm{N=}}\dfrac{{{\rm{no}}{\rm{. \ of \ gram \ equivalent \ of \ solute}}}}{{{\rm{Volume \ of the \ solution \ in \ litres}}}}\].
Complete Step by Step Solution:
Normality is the ratio of no.of gram equivalents of the solute dissolved per litre of a given solution.
If 'w' gram of the solute is present in V litres of a given solution,
\[{\rm{N=}}\left( {\dfrac{{\rm{w}}}{{{\rm{eq}}{\rm{.mass \ of \ the \ solute}}}}} \right){\rm{ \times }}\left( {\dfrac{{{\rm{1000}}}}{{\rm{V}}}} \right)\]
Here, we are given the mass of pure sodium metal, 0.115g.
The volume of the given solution = 500mL.
Equivalent mass of sodium metal in the solution
\[ = \left( {\dfrac{{{\rm{atomic \ mass}}}}{{{\rm{valency}}}}} \right)\]
The atomic mass of Sodium is 23g. The valency of Sodium is 1.
So, the equivalent mass of Sodium
= 23 g/equivalent = 23 amu (atomic mass unit)
So, the normality of Sodium metal in the given solution
$=\left ( \dfrac{0.115 \ g}{23 \ amu} \right )\times\left (\dfrac{1000}{500 \ L} \right )$
$= 0.010 N $
So, option A is correct.
Additional Information:Molarity and Normality comprises volumes of solutions. The temperature has a significant effect on volumes. These values change with temperature. So, molality, mole fraction, mass fraction, etc. have more preference as these depend on masses. Mass does not change with temperature.
Note: While attending to the question, one must have remembered the formula of normality. The equivalent mass of a solute formula also has to be remembered. Units of different values have to be mentioned during calculations.
Formula Used:
\[{\rm{N=}}\dfrac{{{\rm{no}}{\rm{. \ of \ gram \ equivalent \ of \ solute}}}}{{{\rm{Volume \ of the \ solution \ in \ litres}}}}\].
Complete Step by Step Solution:
Normality is the ratio of no.of gram equivalents of the solute dissolved per litre of a given solution.
If 'w' gram of the solute is present in V litres of a given solution,
\[{\rm{N=}}\left( {\dfrac{{\rm{w}}}{{{\rm{eq}}{\rm{.mass \ of \ the \ solute}}}}} \right){\rm{ \times }}\left( {\dfrac{{{\rm{1000}}}}{{\rm{V}}}} \right)\]
Here, we are given the mass of pure sodium metal, 0.115g.
The volume of the given solution = 500mL.
Equivalent mass of sodium metal in the solution
\[ = \left( {\dfrac{{{\rm{atomic \ mass}}}}{{{\rm{valency}}}}} \right)\]
The atomic mass of Sodium is 23g. The valency of Sodium is 1.
So, the equivalent mass of Sodium
= 23 g/equivalent = 23 amu (atomic mass unit)
So, the normality of Sodium metal in the given solution
$=\left ( \dfrac{0.115 \ g}{23 \ amu} \right )\times\left (\dfrac{1000}{500 \ L} \right )$
$= 0.010 N $
So, option A is correct.
Additional Information:Molarity and Normality comprises volumes of solutions. The temperature has a significant effect on volumes. These values change with temperature. So, molality, mole fraction, mass fraction, etc. have more preference as these depend on masses. Mass does not change with temperature.
Note: While attending to the question, one must have remembered the formula of normality. The equivalent mass of a solute formula also has to be remembered. Units of different values have to be mentioned during calculations.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced 2025 Notes
