Answer
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Hint: Equivalent weight of the compound is defined as the molecular weight of the compound per number of equivalent molecules of it. Equivalent moles are also called the valence factor which depends upon the chemical properties of the compound. For example, for acids, the equivalent number of moles is simply no release. For, no equivalent mole is equal to 1.
$\text{ Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$
Complete step by step solution:
Sodium hydroxide has the general formula as$\text{ NaOH }$. We have to find the equivalent weight of Sodium hydroxide. Sodium hydroxide which is shown above has one removable or basic$\text{ O}{{\text{H}}^{-}}\text{ }$. In other words, we can say that one mole of sodium hydroxide requires one mole of monobasic acid .the reaction of sodium hydroxide with an acid such as hydrochloric acid $\text{ HCl }$ is as shown below,
$\text{ }\begin{matrix}
\text{HCl} & \text{+} & \text{NaOH} & \to & \text{NaCl} & + & {{\text{H}}_{\text{2}}}\text{O } \\
1\text{ mole} & {} & 1\text{ mole } & {} & {} & {} & {} \\
\end{matrix}$
On the reaction of sodium hydroxide with acid, it neutralizes the 1 moles of$\text{ O}{{\text{H}}^{-}}\text{ }$.
We are interested to find out the equivalent weight of sodium hydroxide.
We know the formula for equivalent weight as follows,
$\text{ Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$
We have to first calculate the molecular weight of sodium hydroxide $\text{ NaOH }$
The molecular weight of sodium is$\text{ 23 g / mol}$ , of oxygen is $\text{ 16 g / mol}$and for hydrogen is equal to$\text{ 1 g / mol}$.
Let’s calculate the molecular weight of sodium hydroxide,$\text{ Mol}\text{.wt of NaOH = 23 + 16 + 1 = 40 g /mol }$
In sodium hydroxide, there is one hydroxide ion. Thus the no of equivalent moles is 1.
Substitute the values of molecular weight and no of equivalent moles in the equation (1), we get
$\text{ Equivalent weight of NaOH =}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}\text{=}\dfrac{\text{40 g/mol}}{\text{1}}\text{ = 40 g/mol}$
Thus, the equivalent weight of sodium hydroxide $\text{ NaOH }$ is\[\text{40 g/mol}\].
Hence, (A) is the correct option.
Note: In such types of questions, sometimes it can be confusing between moles and the number of equivalent moles. No of moles are given as the ratio of the weight of substance by molecular weight however no of equivalent moles is dependent on the characteristic properties of the substance. This is also called a valence factor.
For acids, valence factor for acid=basicity=removal no.of proton (${{\text{H}}^{\text{+}}}$)
For bases, valence factor for base=acidity=removal no.of hydroxide (\[O{{H}^{-}}\])
$\text{ Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$
Complete step by step solution:
Sodium hydroxide has the general formula as$\text{ NaOH }$. We have to find the equivalent weight of Sodium hydroxide. Sodium hydroxide which is shown above has one removable or basic$\text{ O}{{\text{H}}^{-}}\text{ }$. In other words, we can say that one mole of sodium hydroxide requires one mole of monobasic acid .the reaction of sodium hydroxide with an acid such as hydrochloric acid $\text{ HCl }$ is as shown below,
$\text{ }\begin{matrix}
\text{HCl} & \text{+} & \text{NaOH} & \to & \text{NaCl} & + & {{\text{H}}_{\text{2}}}\text{O } \\
1\text{ mole} & {} & 1\text{ mole } & {} & {} & {} & {} \\
\end{matrix}$
On the reaction of sodium hydroxide with acid, it neutralizes the 1 moles of$\text{ O}{{\text{H}}^{-}}\text{ }$.
We are interested to find out the equivalent weight of sodium hydroxide.
We know the formula for equivalent weight as follows,
$\text{ Equivalent weight=}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}$
We have to first calculate the molecular weight of sodium hydroxide $\text{ NaOH }$
The molecular weight of sodium is$\text{ 23 g / mol}$ , of oxygen is $\text{ 16 g / mol}$and for hydrogen is equal to$\text{ 1 g / mol}$.
Let’s calculate the molecular weight of sodium hydroxide,$\text{ Mol}\text{.wt of NaOH = 23 + 16 + 1 = 40 g /mol }$
In sodium hydroxide, there is one hydroxide ion. Thus the no of equivalent moles is 1.
Substitute the values of molecular weight and no of equivalent moles in the equation (1), we get
$\text{ Equivalent weight of NaOH =}\dfrac{\text{Molecular weight}}{\text{No of equivalent moles}}\text{=}\dfrac{\text{40 g/mol}}{\text{1}}\text{ = 40 g/mol}$
Thus, the equivalent weight of sodium hydroxide $\text{ NaOH }$ is\[\text{40 g/mol}\].
Hence, (A) is the correct option.
Note: In such types of questions, sometimes it can be confusing between moles and the number of equivalent moles. No of moles are given as the ratio of the weight of substance by molecular weight however no of equivalent moles is dependent on the characteristic properties of the substance. This is also called a valence factor.
For acids, valence factor for acid=basicity=removal no.of proton (${{\text{H}}^{\text{+}}}$)
For bases, valence factor for base=acidity=removal no.of hydroxide (\[O{{H}^{-}}\])
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