Question

0.2 mol of HCl and 0.1 mol of barium chloride were dissolved in water to produce a 500 ml solution. The molarity of the Cl is
A. 0.06M
B. 0.09M
C. 0.12M
D. 0.80M

Answer 1

To solve this question, we will first find the number of moles of $C{{l}^{-}}$,that we are getting from HCl and from barium chloride. Then we will find molarity of $C{{l}^{-}}$ by the formula:
$Molarity=\dfrac{Number\text{ }of\text{ }moles\text{ }of\text{ }element}{Volume\text{ }of\text{ }solution\text{ }in\text{ }litres}$

Complete Step by step solution:
- We are getting 0.2moles of $C{{l}^{-}}$,from 0.2 moles of HCl. As:
$0.2\text{ }mole\text{ HCl}\to 0.2\text{ }moles\text{ }of\text{ }C{{l}^{-}}$
- We are getting 0.2 moles of $C{{l}^{-}}$, from 0 .1 moles of barium chloride. As we can see from the reaction:
$CaC{{l}_{2}}\to Ca+2Cl$
Hence,

& 0.1\text{ }mole\text{ }CaC{{l}_{2}}\to 0.1\times 2\text{ }moles\text{ }of\text{ }C{{l}^{-}} \\\ & \Rightarrow 0.2\text{ }moles\text{ }C{{l}^{-}} \\\ \end{aligned}$$ Therefore, we can say that the total moles of $C{{l}^{-}}$is equal to 0.4 moles in 500 ml solution. \- As we know that the formula of Molarity is: $$Molarity=\dfrac{Number\text{ }of\text{ }moles\text{ }of\text{ }element}{Volume\text{ }of\text{ }solution\text{ }in\text{ }litres}$$ \- Now, we can calculate the molarity of $C{{l}^{-}}$ by putting all the values in the above formula: $$\begin{aligned} & Molarity=\dfrac{0.4}{500}\times 1000 \\\ & =0.4\times 2 \\\ & =0.8\text{ }M \\\ \end{aligned}$$ Here, we have converted the volume of solution given in ml into litres. **\- Hence, we can conclude that the correct option is (D), that is the molarity of the Cl is 0.8M.** **Note:** \- We should not get confused in the terms molality and molarity. As, molality is the ratio of the number of moles of solute in kg of solvent. Molality is denoted by m. \- Whereas molarity is the ratio of number of moles of solute in volume of solution. Molarity is denoted by M. \- One should not forget to write the unit in the solution.