Question

TA solution of 10 M NaOH was used to prepare 2L of 0.5 M NaOH, How many mL of the original NaOH solution are needed?
(A) 10mL
(B) 100mL
(C) 1000mL
(D) 200mL
(E) 2000mL

Answer 1

To the original solution water is added to make the new solution of low molarity, and less diluted. Since there is no addition or removal of NaOH in the new and original solution, therefore, moles of NaOH will be the same in both the solutions.

Complete step by step solution:
Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litre of a solution. Molarity is also known as the molar concentration of a solution.
The formula of molarity is as follows:
$molarity=\dfrac{\text{no}\text{. of moles}}{volume(L)}$
In the given, we are given information about two solutions, one is 10 M NaOH whose volume is not known, and other is 2L of 0.5 M NaOH. 0.5 M NaOH is prepared using 10 M NaOH solution. Since the newly formed solution is more diluted than the original solution, therefore, water must be added to the original solution for dilution. Since the number of moles of NaOH is equal in both the solutions therefore, we can use the above formula to find out the volume of the original solution. Therefore, the relation will be,
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
Where, ${{M}_{1}}$ is the molarity of the original solution,
${{V}_{1}}$ is the volume of the original solution,
${{M}_{2}}$ is the molarity of a formed solution,
${{V}_{2}}$ is the volume of the formed solution.
${{V}_{1}}=\dfrac{{{M}_{2}}{{V}_{2}}}{{{M}_{1}}}$
${{V}_{1}}=\dfrac{2\times 0.5}{10}=0.1L=100mL$
Therefore, 100mL of water is added to the original solution.

Hence the correct answer is the (B) option.

Note: Remember that as the options are mentioned in mL, make sure that you consider the volume in mL only. Sodium hydroxide is used to manufacture soaps, rayon, paper, explosives, dyestuffs, and petroleum products. It is also used in processing cotton fabric, laundering and bleaching, metal cleaning and processing, oxide coating, electroplating, and electrolytic extracting.