
Normality of solution 20.5 gm of NaOH solution in 1000 mL of solution is:
Answer
483.9k+ views
Hint: Normality formula will be almost the same as that of molarity. But we take the number of equivalents for calculating normality whereas we take the number of moles for calculating molarity. Try to figure out what equivalents mean.
Complete step by step answer:
We are asked to calculate the normality of NaOH in the given solution. We know that the normality of a substance is the number of equivalents of substance present in a 1L volume of solution. Equivalents calculation will be different for different types of substances.
For bases equivalent weight = $\frac { molecular\quad weight\quad of\quad base }{ acidity\quad of\quad base } $
The acidity of a base means the number of ionizable hydroxide groups in the base.
NaOH has one ionizable hydroxide group.
The equivalent weight of NaOH = $\frac { molecular\quad weight\quad of\quad NaOH }{ acidity\quad of\quad NaOH } \quad =\quad \frac { 40 }{ 1 } \quad =\quad 40$
Normality of NaOH solution = $\frac { Weight\quad of\quad NaOH }{ equivalent\quad weight\quad of\quad NaOH } \times \frac { 1 }{ volume\quad of\quad solution\quad in\quad liters } $
Volume of solution = 1L(1000mL)
Weight of NaOH = 20.5 gm
Normality of NaOH = $\frac { 20.5 }{ 40 } \times \frac { 1 }{ 1 } \quad =\quad \frac { 20.5 }{ 40 } \quad =\quad \frac { 41 }{ 80 } \quad =\quad 0.5125N$
Therefore, the normality of NaOH is 0.5125N.
Note:
In this problem, normality is the same as that of molarity but in some cases where the equivalent weight will not be equal to molecular weight we need to calculate the correct number of equivalents. In redox reactions equivalent weight of a substance that undergoes oxidation or reduction will be the molecular weight of substance divided by the number of electrons transferred by one mole of the substance.
Complete step by step answer:
We are asked to calculate the normality of NaOH in the given solution. We know that the normality of a substance is the number of equivalents of substance present in a 1L volume of solution. Equivalents calculation will be different for different types of substances.
For bases equivalent weight = $\frac { molecular\quad weight\quad of\quad base }{ acidity\quad of\quad base } $
The acidity of a base means the number of ionizable hydroxide groups in the base.
NaOH has one ionizable hydroxide group.
The equivalent weight of NaOH = $\frac { molecular\quad weight\quad of\quad NaOH }{ acidity\quad of\quad NaOH } \quad =\quad \frac { 40 }{ 1 } \quad =\quad 40$
Normality of NaOH solution = $\frac { Weight\quad of\quad NaOH }{ equivalent\quad weight\quad of\quad NaOH } \times \frac { 1 }{ volume\quad of\quad solution\quad in\quad liters } $
Volume of solution = 1L(1000mL)
Weight of NaOH = 20.5 gm
Normality of NaOH = $\frac { 20.5 }{ 40 } \times \frac { 1 }{ 1 } \quad =\quad \frac { 20.5 }{ 40 } \quad =\quad \frac { 41 }{ 80 } \quad =\quad 0.5125N$
Therefore, the normality of NaOH is 0.5125N.
Note:
In this problem, normality is the same as that of molarity but in some cases where the equivalent weight will not be equal to molecular weight we need to calculate the correct number of equivalents. In redox reactions equivalent weight of a substance that undergoes oxidation or reduction will be the molecular weight of substance divided by the number of electrons transferred by one mole of the substance.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Find the image of the point 38 about the line x+3y class 11 maths CBSE

Can anyone list 10 advantages and disadvantages of friction

Distinguish between Mitosis and Meiosis class 11 biology CBSE
