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Question: The amount of \[BaC{{l}_{2}}\](in g) needed to make \(250\)ml of a solution having the same concentr...

The amount of BaCl2BaC{{l}_{2}}(in g) needed to make 250250ml of a solution having the same concentration of ClC{{l}^{-}}as in the one containing 3.783.78g of NaClNaClper 100100ml is (multiply the answer by 1010)

Explanation

Solution

Two solutions of the same concentration means the two solutions given in the question will have the same molarity value. So, we can get the answer by calculating the molarity of NaClNaClfirst using a direct formula of molarity and then similarly will apply for BaCl2BaC{{l}_{2}}to find its mass.

Complete step by step solution:
As per the given question,
The concentration of the solution containing BaCl2BaC{{l}_{2}}is equal to the concentration of the solution containing NaClNaCl. That means, two of the solutions will have the same molarity value. Therefore, first we have to calculate the molarity of NaClNaClsolution.
Given that,
Mass of NaClNaClis 3.783.78g.
Volume of NaClNaClsolution is 100100ml i.e. 100100per 10001000litre.
Volume of BaCl2BaC{{l}_{2}}solution is 250250ml i.e. 250250per 10001000litre.
Mass of BaCl2BaC{{l}_{2}}is to be calculated.
So, let’s consider the mass of BaCl2BaC{{l}_{2}}be ‘X’ g.
As we know,
Molar mass of NaClNaCl is58.5g/mol58.5g/mol.
And, formula of molarity is:
Molarity=Mass(g)MolarMass(g/mol)×VolumeSolution(l)Molarity=\dfrac{Mass(g)}{Molar Mass(g/mol)\times Volume Solution(l)}
So, Molarity of NaClNaClwill be
MolarityofNaCl=3.78×100058.5×100=0.65mol/lMolarityofNaCl=\dfrac{3.78\times 1000}{58.5\times 100}=0.65mol/l
Now, we know the molarity of NaClNaClis equal to BaCl2BaC{{l}_{2}}.
So, MolarityBaCl2=massBaCl2ingramsmolarmassofBaCl2×VolumeOfSolutionInLitresMolarity BaC{{l}_{2}}=\dfrac{mass BaC{{l}_{2}}ingrams}{molarmassofBaC{{l}_{2}}\times Volume Of Solution In Litres}
Mass of BaCl2BaC{{l}_{2}}is X g.
Molar mass of BaCl2BaC{{l}_{2}}is 208.33g/mol208.33g/mol.
Volume of BaCl2BaC{{l}_{2}}solution is 250250ml i.e. 250250per 10001000litre.
So, molarity of BaCl2BaC{{l}_{2}}will be,
MolarityBaCl2=X×1000208.33×250=4X208.33g/molMolarity BaC{{l}_{2}}=\dfrac{X\times 1000}{208.33\times 250}=\dfrac{4X}{208.33}g/mol
And, the molarity of BaCl2BaC{{l}_{2}}is the same as that of NaClNaCl.
So, 4X208.33=0.65\dfrac{4X}{208.33}=0.65
Then,
X=0.65×208.334X=\dfrac{0.65\times 208.33}{4}
Thus, X=33.8X=33.8
So, the mass of BaCl2BaC{{l}_{2}}is 33.8g33.8g.
As per the given question, we have to multiply the result with 1010.
So, the mass of BaCl2BaC{{l}_{2}}will be 3.38g3.38g.

Hence, 3.38g3.38g of BaCl2BaC{{l}_{2}} is required to make 250250ml of solution having same concentration of chloride as one containing 3.783.78g of NaClNaCl per ml.

Note: Molarity indicates the number of moles of solute per litre of a solution and is one of the most common units used to measure the concentration of a solution. It can also be used to calculate the volume of a solvent or the amount of solute. So, when it is said two solutions have equal concentration, it means that they have the same molar concentration.