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# Which of the following should be done in order to prepare 0.40M NaCl starting with 100ml of 0.30 M NaCl? (Mol. Wt. of NaCl= 58.5)(A) Add 0.585 g of NaCl(B) Add 20 ml of water(C) Add 0.010 ml NaCl(D) Evaporate 10 ml of water

Last updated date: 02nd Aug 2024
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Hint: When you start from the lower concentration of the solution and want to prepare a higher concentrated solution from the same, determine the number of moles required to prepare the solutions. Number of moles can be determined by the molarity. It is as shown below,
$\text{ Molarity = }\dfrac{\text{No}\text{.of moles }}{\text{Volume in d}{{\text{m}}^{\text{3}}}}\text{ = }\dfrac{\text{weight}}{\text{molar mass}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{1 d}{{\text{m}}^{\text{3}}}}\text{ }$
Determine the moles and then the amount of solute which is needed to be added to the existing solution.

Complete step by step solution:
We have been provided with a molecular weight of NaCl that is 58.5,
We need to prepare 0⋅4M NaCl solution and we are starting with a 100 ml, 0⋅3M NaCl solution.
So, for that:
We will be firstly finding the moles for 100ml of 0.30M, by using the formula: mole= given mass/ molar mass,
But as we are not provided with mass,
So, we need to determine it first, by using: $mass=volume\times molarity$,
So, for 100ml of 0.30M: $mass = 100\times 0.3$,
Now keeping this value in mole = given mass/ molar mass,
So, moles would be: $moles=\dfrac{100\times 0.3}{1000} = 0.03mole$,
- Now, for 100ml of 0.40M also firstly we need to find the mass using the formula of: $mass=volume\times molarity$,
So, it would come out to be: $mass = 100\times 0.4$,

- Now keeping this value in mole= given mass/ molar mass,
So, the moles would be:
$moles=\dfrac{100\times 0.4}{1000} = 0.04mole$,
So, the moles of NaCl to be added would be 0.04 - 0.03 = 0.01 mole,
Which would be: $0.01\times 58.5g = 0.585g$,
So, we can say that we need to add 0.585g of NaCl in order to prepare 0.40M NaCl starting with 100ml of 0.30 M NaCl,
So, the correct answer is “Option A”.

Note: It is a deceiving question. We know the relation between the molarities and volume i.e.$\text{ }{{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\text{ = }{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}\text{ }$ .student would tempt to solve this question by applying this relation. But have a second look at the question. We are asked to prepare a higher concentration from the lower .Thus we will add extra solute to the existing solution.