Question

The amount of \[BaC{{l}_{2}}\](in g) needed to make $250$ml of a solution having the same concentration of \[C{{l}^{-}}\]as in the one containing $3.78$g of \[NaCl\]per $100$ml is (multiply the answer by $10$)

Answer 1

Hint: 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 \[NaCl\]first using a direct formula of molarity and then similarly will apply for \[BaC{{l}_{2}}\]to find its mass.


Complete step by step solution:
As per the given question,
The concentration of the solution containing \[BaC{{l}_{2}}\]is equal to the concentration of the solution containing \[NaCl\]. That means, two of the solutions will have the same molarity value. Therefore, first we have to calculate the molarity of \[NaCl\]solution.
Given that,
Mass of \[NaCl\]is $3.78$g.
Volume of \[NaCl\]solution is $100$ml i.e. $100$per $1000$litre.
Volume of \[BaC{{l}_{2}}\]solution is $250$ml i.e. $250$per $1000$litre.
Mass of \[BaC{{l}_{2}}\]is to be calculated.
So, let’s consider the mass of \[BaC{{l}_{2}}\]be ‘X’ g.
As we know,
Molar mass of \[NaCl\] is$58.5g/mol$.
And, formula of molarity is:
$Molarity=\dfrac{Mass(g)}{Molar Mass(g/mol)\times Volume Solution(l)}$
So, Molarity of \[NaCl\]will be
$MolarityofNaCl=\dfrac{3.78\times 1000}{58.5\times 100}=0.65mol/l$
Now, we know the molarity of \[NaCl\]is equal to \[BaC{{l}_{2}}\].
So, \[Molarity BaC{{l}_{2}}=\dfrac{mass BaC{{l}_{2}}ingrams}{molarmassofBaC{{l}_{2}}\times Volume Of Solution In Litres}\]
Mass of \[BaC{{l}_{2}}\]is X g.
Molar mass of \[BaC{{l}_{2}}\]is $208.33g/mol$.
Volume of \[BaC{{l}_{2}}\]solution is $250$ml i.e. $250$per $1000$litre.
So, molarity of \[BaC{{l}_{2}}\]will be,
\[Molarity BaC{{l}_{2}}=\dfrac{X\times 1000}{208.33\times 250}=\dfrac{4X}{208.33}g/mol\]
And, the molarity of \[BaC{{l}_{2}}\]is the same as that of \[NaCl\].
So, \[\dfrac{4X}{208.33}=0.65\]
Then,
\[X=\dfrac{0.65\times 208.33}{4}\]
Thus, \[X=33.8\]
So, the mass of \[BaC{{l}_{2}}\]is $33.8g$.
As per the given question, we have to multiply the result with $10$.
So, the mass of \[BaC{{l}_{2}}\]will be $3.38g$.

Hence, $3.38g$ of \[BaC{{l}_{2}}\] is required to make $250$ml of solution having same concentration of chloride as one containing $3.78$g of \[NaCl\] 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.