
If 5.85 g of NaCl (molecular weight = 58.5) is dissolved in water and the solution is made up to 0.5 litre, the molarity of the solution will be:
(a) 0.2
(b) 0.4
(c) 1.0
(d) 0.1
Answer
223.5k+ views
Hint: Firstly, calculate the number of moles by dividing weight of the compound by its molecular weight. Then, use this to calculate the molarity of the solution in moles per litre volume.
Complete step by step answer:
Molarity, often represented by ‘M’, is a term used to express concentration in terms of moles per litre of a solution.
Therefore, we define it as - “Molarity of any solution is number of moles of solute per litre of solution”. It has the following formula –
\[\text{Molarity=}\dfrac{\text{moles of solute (mol)}}{\text{volume of solution (L)}}\]
From the above formula, we can see that the SI unit of Molarity is mol/L.
According to the question,
weight of NaCl = 5.85 g
molecular weight of NaCl= 58.5
From this, we can calculate the number of moles of NaCl
= \[\dfrac{\text{weight of NaCl}}{\text{molecular weight of NaCl}}\]
= \[\dfrac{5.85}{58.5}\] = 0.1 moles
Now, let us calculate the Molarity of solution –
Molarity = \[\dfrac{\text{moles of solute (mol)}}{\text{volume of solution (L)}}=\dfrac{0.1}{0.5}\]mol/L = 0.2 mol/L
Therefore, the answer is – option (a) – the molarity of the solution is 0.2 mol/L.
Additional Information:
The ‘Concentration’ of a solution is defined as the relative amount of solute present in a solution.
Note: People often confuse between Molarity and Molality. Molality is represented by ‘m’, whereas molarity is represented by ‘M’. Molality of any solution can be defined as the number of moles of solute present per kg of solvent. It can be represented by the following –
\[\text{Molality=}\dfrac{\text{moles of solute (mol)}}{\text{weight of solvent (kg)}}\]
Its SI unit is mol/kg.
Complete step by step answer:
Molarity, often represented by ‘M’, is a term used to express concentration in terms of moles per litre of a solution.
Therefore, we define it as - “Molarity of any solution is number of moles of solute per litre of solution”. It has the following formula –
\[\text{Molarity=}\dfrac{\text{moles of solute (mol)}}{\text{volume of solution (L)}}\]
From the above formula, we can see that the SI unit of Molarity is mol/L.
According to the question,
weight of NaCl = 5.85 g
molecular weight of NaCl= 58.5
From this, we can calculate the number of moles of NaCl
= \[\dfrac{\text{weight of NaCl}}{\text{molecular weight of NaCl}}\]
= \[\dfrac{5.85}{58.5}\] = 0.1 moles
Now, let us calculate the Molarity of solution –
Molarity = \[\dfrac{\text{moles of solute (mol)}}{\text{volume of solution (L)}}=\dfrac{0.1}{0.5}\]mol/L = 0.2 mol/L
Therefore, the answer is – option (a) – the molarity of the solution is 0.2 mol/L.
Additional Information:
The ‘Concentration’ of a solution is defined as the relative amount of solute present in a solution.
Note: People often confuse between Molarity and Molality. Molality is represented by ‘m’, whereas molarity is represented by ‘M’. Molality of any solution can be defined as the number of moles of solute present per kg of solvent. It can be represented by the following –
\[\text{Molality=}\dfrac{\text{moles of solute (mol)}}{\text{weight of solvent (kg)}}\]
Its SI unit is mol/kg.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: City Intimation Slip Releasing Today, Application Form Closed, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Redox Reaction Class 11 Chemistry Chapter 7 CBSE Notes - 2025-26

