Here are few noteworthy points to keep in mind concerning this topic📋

**EQUATION 1**

**Molarity (mol dm ^{-3}) = moles of solute (mol) **

**÷ volume of soltion (dm**

^{3})**1 dm ^{3} = 1000 c**

**m**

^{3}🧐**Pay attention to the units**❗️ ❕

**EQUATION 2**

**Moles = Mass ****÷ **** Molar Mass**

Time for the examples❗️ ❕

**Example 1**🤓

**5.78 g of CaCl _{2} are dissolved in water and the solution is made up to 250 cm^{3}. Calculate the molarity of the solution.**

**(Relative atomic masses: Cl= 35.5, Ca= 40.1)**

**Step 1- Find the number of moles of CaCl _{2}**

**Molar Mass of CaCl _{2} = (40.1+35.5+35.5) = 111.1 g mol^{-1}**

Moles of CaCl_{2}= 5.78 g ÷ 111.1 g mol^{-1} = 0.0520 mol

**Final Step- Find the molarity by substituting the correct values in EQUATION 1**

**Volume = 250 ÷ 1000 = 0.25**

Molarity= 0.0520 ÷ 0.25 = 0.208 M

**Example 2**🤓

**What is the mass of Na _{2}SO_{4} required to prepare 1.5 dm^{3} of 1.6 mol **

**dm**

^{-3}?

**(Relative atomic masses: O=16, Na= 32.1, S=32.1)**

**It can be seen that the question has provided the volume and molarity.**

**So the number of moles can be found**

**Step 1- Find the moles of** **Na _{2}SO_{4} by REARRANGING **

**EQUATION 1**

Moles of Na_{2}SO_{4} = Molarity** **× Volume = 1.5 × 1.6 = 2.4 mol

**Last Step- Find the mass of Na _{2}SO_{4 }by REARRANGING **

**EQUATION 2**

**Molar Mass of Na _{2}SO_{4} = 142.1 g mol^{-1}**

Mass of Na_{2}SO_{4} = Moles × Molar Mass = 2.4 × 142.1 = 341 g

**Example 3**🤓

**Calculate the amount, in moles, of 25cm ^{3} of hydrochloric acid (HCl) with a concentration of 2 mol dm^{3}**

**Step 1- Convert to the correct unit**

**Volume = 25 ÷ 1000 = 0.025**

**Final Step-** **Find the moles of HCl by REARRANGING ****EQUATION 1**

Moles of HCl = Molarity** **× Volume = 2 × 0.025 = 0.05

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