Question

If \[{\text{0}}{\text{.45g}}\] of acid (molecular weight = 90) was exactly neutralized by \[20{\text{mL}}\] of \[0.5{\text{N}}\]\[{\text{NaOH}}\]. Basicity of the acid is:
A.1
B.2
C.3
D.4

Answer 1

Hint:To answer this question, you should recall the concept of normality and neutralization reaction. When equal equivalents of acid and base are mixed it leads to a neutral solution. We shall substitute the values given in the given formula.
Formula used:
${\text{Normality = Molarity}} \times {\text{n - factor}}$

Complete step by step answer:
Let the \[{n_{factor}}\] ​ be \[n.\]
Moles of acid \[ = 900.45{\text{ }} = 5{\text{millimoles}}\]
Comparing equivalents of acid and NaOH
\[5 \times n = 0.5 \times 20\].
\[ \Rightarrow n = 2\]
For acid, \[{n_{factor}} = \] basicity.
$\therefore $ Basicity is 2.

Hence, the correct option is B.
Note:
You should know about the other concentration terms commonly used:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (\[{10^6}\]) of the solution.
\[{\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}\]
Normality: It is defined as the number of gram equivalents of solute present in one litre of the solution.
Mole Fraction: It gives a unitless value and is defined as the ratio of moles of one component to the total moles present in the solution.
Mole fraction = $\dfrac{{{X_{\text{A}}}}}{{{X_{\text{A}}} + {X_B}}}$(from the above definition) where ${X_{\text{A}}}$is no. of moles of glucose and \[{X_{\text{B}}}\]is the no. of moles of solvent.