Find the equivalent weight of calcium hydroxide? $\left( {Ca = 40,O = 16,H = 1} \right)$
(A) $23u$
(B) $37u$
(C) $32u$
(D) $56u$
Answer
Verified
468k+ views
Hint: Equivalent weight is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance. The equivalent weight of an element is the mass which combines with or displaces $1.008grams$ of hydrogen, or $35.5grams$ of chlorine.
Complete step by step answer:
The molecular formula of calcium hydroxide is \[Ca{\left( {OH} \right)_2}\]
Molecular mass of $Ca{\left( {OH} \right)_2} = 40 + \left( {16 + 1} \right) \times 2$
$ = 40 + \left( {17 \times 2} \right)$
$ = 74u$
The formula/expression for the equivalent weight of $Ca{\left( {OH} \right)_2}$ is given as follows:
$Equivalent{\text{ }}weight = \dfrac{{molecular{\text{ }}mass}}{{no.{\text{ }}of{\text{ }}replaceable{\text{ }}O{H^ - }ions}}$
$Ca{\left( {OH} \right)_2} \rightleftharpoons C{a^{2 + }} + 2O{H^ - }$
From this reaction, we can say that there are $2$ replaceable $O{H^ - }$ groups.
Substitute the values in the above expression:
Equivalent weight $ = \dfrac{{74u}}{2} = 37u$
So, the correct answer is Option B.
Note:
Equivalent weight unlike molecular weight is proportional mass of chemical entities which combine or displace other chemical entities.
The idea of equivalent mass to compare chemically different elements. Atoms combine with each other to form chemical compounds, such that the elements are always present in definite proportions by mass and this property can be used to make chemically different molecules with different mass, equal.
Equivalent mass is determined experimentally and has dimensions and units of mass. Equivalent mass concept is of greater importance in analytical chemistry because in order to follow primary standards the compounds with greater equivalent mass are preferred in order to reduce errors in calculation.
It is also widely used in polymer chemistry during ion-exchange reaction where one equivalent of ion exchange polymer will exchange with one mole of singly charged ions.
Equivalent weight =$\dfrac{{molecular{\text{ }}weight}}{{N{\text{ }}factor}}$
Normality can be find out form the equivalent weight form the following formula:
$Normality = \dfrac{{no.{\text{ }}of{\text{ }}equivalent{\text{ }}weight}}{{litres{\text{ }}of{\text{ }}solution}}$
Complete step by step answer:
The molecular formula of calcium hydroxide is \[Ca{\left( {OH} \right)_2}\]
Molecular mass of $Ca{\left( {OH} \right)_2} = 40 + \left( {16 + 1} \right) \times 2$
$ = 40 + \left( {17 \times 2} \right)$
$ = 74u$
The formula/expression for the equivalent weight of $Ca{\left( {OH} \right)_2}$ is given as follows:
$Equivalent{\text{ }}weight = \dfrac{{molecular{\text{ }}mass}}{{no.{\text{ }}of{\text{ }}replaceable{\text{ }}O{H^ - }ions}}$
$Ca{\left( {OH} \right)_2} \rightleftharpoons C{a^{2 + }} + 2O{H^ - }$
From this reaction, we can say that there are $2$ replaceable $O{H^ - }$ groups.
Substitute the values in the above expression:
Equivalent weight $ = \dfrac{{74u}}{2} = 37u$
So, the correct answer is Option B.
Note:
Equivalent weight unlike molecular weight is proportional mass of chemical entities which combine or displace other chemical entities.
The idea of equivalent mass to compare chemically different elements. Atoms combine with each other to form chemical compounds, such that the elements are always present in definite proportions by mass and this property can be used to make chemically different molecules with different mass, equal.
Equivalent mass is determined experimentally and has dimensions and units of mass. Equivalent mass concept is of greater importance in analytical chemistry because in order to follow primary standards the compounds with greater equivalent mass are preferred in order to reduce errors in calculation.
It is also widely used in polymer chemistry during ion-exchange reaction where one equivalent of ion exchange polymer will exchange with one mole of singly charged ions.
Equivalent weight =$\dfrac{{molecular{\text{ }}weight}}{{N{\text{ }}factor}}$
Normality can be find out form the equivalent weight form the following formula:
$Normality = \dfrac{{no.{\text{ }}of{\text{ }}equivalent{\text{ }}weight}}{{litres{\text{ }}of{\text{ }}solution}}$
Recently Updated Pages
How to find how many moles are in an ion I am given class 11 chemistry CBSE
Class 11 Question and Answer - Your Ultimate Solutions Guide
Identify how many lines of symmetry drawn are there class 8 maths CBSE
State true or false If two lines intersect and if one class 8 maths CBSE
Tina had 20m 5cm long cloth She cuts 4m 50cm lengt-class-8-maths-CBSE
Which sentence is punctuated correctly A Always ask class 8 english CBSE
Trending doubts
The reservoir of dam is called Govind Sagar A Jayakwadi class 11 social science CBSE
What problem did Carter face when he reached the mummy class 11 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
Petromyzon belongs to class A Osteichthyes B Chondrichthyes class 11 biology CBSE
Comparative account of the alimentary canal and digestive class 11 biology CBSE