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

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