Question

The molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml solution is
A . 3.05 M
B . 1.35 M
C . 2.50 M
D . 4.50 M

Answer 1

Hint: The concept of "molarity" must be used in this question. First, by dividing the weight of compound KOH by its molecular weight, we must determine the number of moles. The solution's molarity in moles per litre volume can then be determined using this information.

Complete answer:The number of moles of solute present in a specific number of litres of the solution, or moles per litre of a solution, is known as molar concentration or molarity. It is one of the most often used concentration units and is represented by the letter M.
 The formula for it is:
$Molarity=\dfrac{moles\text{ }of\text{ }solute}{volume\text{ }of\text{ }solution\text{ }in\text{ }litres}$
The SI unit for molarity is mol/L.
In the given question, the solute is KOH.
Molar mass of KOH = molar mass of K + molar mass of O + molar mass of H = 39+16+1 = 56g/mol
\[Number\text{ }of\text{ }moles=\dfrac{Given\text{ }mass}{Molar\text{ }mass}\]
Given the mass of KOH = 75.5g
Moles of KOH = $\dfrac{75.5}{56}=1.35$
Volume of solution= 540mL= 0.54L
Therefore, molarity of solution = $\dfrac{1.35}{0.54}=2.5$M

The correct answer is C.

Note: The terms molarity, molality, and normality could all be confused. In contrast to molarity, which is symbolised by the letter "M," molality is represented by the letter "m." The molality of any solution is defined as the moles of solute per kilogram of solvent. It can be expressed by:
$Molality=\dfrac{moles\text{ }per\text{ }solute\left( mol \right)}{weight\text{ }of\text{ }solvent\text{ }in\text{ }kg}$
It has the SI unit mol/kg.
The term "normality" refers to the quantity of solute in a solution measured in grams or moles per litre. It is commonly referred to as a solution's equivalent concentration and is denoted by the letter "N". N or equivalents/Litre serve as the units of normality.