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# 50mL of 20.8 % w/v $BaC{{l}_{2}}$(aq) and 100mL of 9.8 % w/v ${{H}_{2}}S{{O}_{4}}$(aq) solutions were mixed. Molarity of $C{{l}^{-}}$ ion in the resulting solution is? (At. Mass of Ba=137)A. 0.333 MB. 0.666 MC. 0.1 MD. 1.33 M

Last updated date: 22nd Jun 2024
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Hint:. Try to find an equation that relates the molarity with the number of moles and the volume given of the solution. Volume of the solution is already given in the question, only a number of moles is needed to be found using the given data.

- In order to answer our question, we need to learn about the number of moles and molarity. In order to solve this question, we need to have an idea about the moles and molar mass of an element or compound. Before the advent of mass spectrometry for determining the atomic masses accurately, scientists were determining mass of one atom relative to another by experimental means. Hydrogen, being lightest atom arbitrarily assigned a mass of 1 (without any units) and other elements were assigned masses relative But in 1961, the International Union of chemists had chosen the most stable isotope of carbon (C-12 isotope) as the default mass for comparison of the atomic masses of various elements. In this system $^{12}C$ is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. Now, the mass of the substance, be it any element or a compound which has an exact number of $6\times {{10}^{23}}$ particles in it is called the molar mass. Molar mass is different for different elements and compounds.
$Molarity(M) = \dfrac{No\,of\,mole{{s}_{solute}}}{Volum{{e}_{solution}}(litre)}$
Now, assuming 100 mL of aqueous solution, we have 20.8 g $BaC{{l}_{2}}$ and 9.8g ${{H}_{2}}S{{O}_{4}}$. So, $\dfrac{20.8}{2} = 10.4g\,BaC{{l}_{2}}$ is present in 50 mL of the solution, and mass of $C{{l}^{-}}$ is $\dfrac{35.5}{10} = 3.55g$
After mixing, the total volume of the solution becomes $50 + 100 = 150mL$. Mass of $C{{l}^{-}}$ will remain same and we obtain the number of moles as:
$no\,of\,mole{{s}_{C{{l}^{-}}}}=\dfrac{3.55g}{35.5g\,mo{{l}^{-1}}} = 0.1\,moles$
$molarit{{y}_{C{{l}^{-}}}}=\dfrac{0.1\times 1000}{150} = 0.66M$
So, we obtain the molarity of $C{{l}^{-}}$ ion as 0.66M So, the correct answer is “Option B”.