Question

The volume of water that would convert \[{\text{10ml}}\] of decimolar \[{\text{HCl}}\] solution to equimolar solution is:
A.\[{\text{1}}{{\text{0}}^{\text{3}}}{\text{ml}}\]
B.\[{\text{1}}{{\text{0}}^{\text{2}}}{\text{ml}}\]
C.\[{\text{9}}{\text{.9}} \times {\text{1}}{{\text{0}}^{\text{2}}}{\text{ml}}\]
D.\[{\text{1}}{\text{.2ml}}\]

Answer 1

Hint: To answer this question, you should recall the concept of molarity. Molarity is defined as the moles of a solute per litre of a solution. We shall use initial molarity to calculate the number of moles of \[{\text{HCl}}\]. Then, we shall find the volume of the solution and thus, the volume of water required.
The formula used: ${\text{Molarity = }}\dfrac{{{\text{no}}{\text{. of moles}}}}{{{\text{Volume}}}}$

Complete step by step answer:
On dilution, the moles \[{\text{HCl}}\] will remain constant.
Initially, we can use the formula of molarity to calculate the number of moles:
\[10 = \dfrac{{{\text{no}}{\text{. of mili moles}}}}{{10}}\]
\[ \Rightarrow {\text{no}}{\text{. of mili moles}} = 100\]
Now, the new molarity is ${\text{0}}{\text{.1M}}$. On dilution the moles of \[{\text{HCl}}\] will remain constant:
\[0.1 = \dfrac{{{\text{no}}{\text{. of mili moles}}}}{{{\text{Volume}}}}\]
Substituting the number of milli-moles
\[ \Rightarrow 0.1 = \dfrac{{100}}{{{\text{Volume}}}}\]
\[ \Rightarrow {\text{Volume}} = 1000{\text{ml}}\]
Thus \[990{\text{ml}}\] of water should be added to \[{\text{10ml}}\] on concentrated \[{\text{HCl}}\] to get a decimolar solution.

Hence, the correct answer to this question is option A.

Note:
Other concentration terms used are:
Concentration in Parts Per Million (ppm): The parts of a component per million parts (\[{10^6}\]) of the solution.
\[{\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total Mass Of The Solution}}}} \times {10^6}\]
Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- \[{\text{Molality(m) = }}\dfrac{{{\text{Moles Of Solute}}}}{{{\text{Mass Of Solvent In Kg}}}}\]
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