
The mass of substance containing 60% NaCl, 37% KCl, that should be weight out for analysis so that, after the action of 25 mL of 0.1N $AgNO_{ 3 }$ solution, excess of $Ag^{ + }$ is back titrated with 5 mL of $NH_{ 4 }SCN$ solution, is …………….. Given that 1 mL of $NH_{ 4 }SCN$ = 1.1 mL of $AgNO_{ 3 }$, write it in multiples of 100 (only two digits).
Answer
243.3k+ views
Hint: This question needs to be solved stepwise. First, calculate the moles of$AgNO_{ 3 }$ and then find the number of moles which will react with NaCl/KCl in their percentage ratio. Finally, calculate the value of the amount of substance for 100% in grams.
Complete step by step answer:
We are given in the question that, 1 mL of $NH_{ 4 }SCN$ = 1.1 mL of $AgNO_{ 3 }$
So, 5 ml of $NH_{ 4 }SCN$ will require 5.5 ml of $AgNO_{ 3 }$. Hence, the volume of $AgNO_{ 3 }$ used in the reaction with the mixture NaCl + KCl will be = 25 - 5.5 = 19.5 ml.
Moles of $AgNO_{ 3 }$ in 0.1N 19.5 ml solution =
$\dfrac { 0.1 }{ 1000 } \quad \times \quad 19.5$ = 0.00195 moles
In the reaction
$NaCl/KCl\quad +\quad AgNO_{ 3 }\quad \rightarrow \quad AgNO_{ 3 }/KNO_{ 3 }$
1 mole of $AgNO_{ 3 }$ reacts with 1 mole of NaCl / KCl in their percentage ratio, hence 0.00195 moles will react with
= $(\dfrac { 60 }{ 100 } \quad \times \quad 0.00195)_{ NaCl }\quad +\quad (\dfrac { 37 }{ 100 } \quad \times \quad 0.00195)_{ KCl }$
= 0.00117 moles of NaCl + 0.0007215 moles of KCl
So, the weight will be
= (0.00117 x 58.5) + (0.0007215 x 74.5)
= 0.1224 g
The 97% of the amount of substance = 0.1224 g
So, 100% will be equal to = (100/97) x 0.1224 = 0.126 g
We need to write this in multiple of 100 and in two digits so it would be 0.13 g.
Therefore, we calculated the mass of the substance which is equal to 0.13 g.
Note: We should also know some important uses of $NH_{ 4 }SCN$.
Ammonium thiocyanate is used in the manufacture of herbicides, thiourea, and transparent artificial resins; in matches; as a stabilizing agent in photography; in various rustproofing compositions; as an adjuvant in textile dyeing and printing; as a tracer in oil fields; in the separation of hafnium from zirconium, and in titrimetric analysis.
Ammonium thiocyanate can also be used to determine the iron content in soft drinks by colorimetry.
Complete step by step answer:
We are given in the question that, 1 mL of $NH_{ 4 }SCN$ = 1.1 mL of $AgNO_{ 3 }$
So, 5 ml of $NH_{ 4 }SCN$ will require 5.5 ml of $AgNO_{ 3 }$. Hence, the volume of $AgNO_{ 3 }$ used in the reaction with the mixture NaCl + KCl will be = 25 - 5.5 = 19.5 ml.
Moles of $AgNO_{ 3 }$ in 0.1N 19.5 ml solution =
$\dfrac { 0.1 }{ 1000 } \quad \times \quad 19.5$ = 0.00195 moles
In the reaction
$NaCl/KCl\quad +\quad AgNO_{ 3 }\quad \rightarrow \quad AgNO_{ 3 }/KNO_{ 3 }$
1 mole of $AgNO_{ 3 }$ reacts with 1 mole of NaCl / KCl in their percentage ratio, hence 0.00195 moles will react with
= $(\dfrac { 60 }{ 100 } \quad \times \quad 0.00195)_{ NaCl }\quad +\quad (\dfrac { 37 }{ 100 } \quad \times \quad 0.00195)_{ KCl }$
= 0.00117 moles of NaCl + 0.0007215 moles of KCl
So, the weight will be
= (0.00117 x 58.5) + (0.0007215 x 74.5)
= 0.1224 g
The 97% of the amount of substance = 0.1224 g
So, 100% will be equal to = (100/97) x 0.1224 = 0.126 g
We need to write this in multiple of 100 and in two digits so it would be 0.13 g.
Therefore, we calculated the mass of the substance which is equal to 0.13 g.
Note: We should also know some important uses of $NH_{ 4 }SCN$.
Ammonium thiocyanate is used in the manufacture of herbicides, thiourea, and transparent artificial resins; in matches; as a stabilizing agent in photography; in various rustproofing compositions; as an adjuvant in textile dyeing and printing; as a tracer in oil fields; in the separation of hafnium from zirconium, and in titrimetric analysis.
Ammonium thiocyanate can also be used to determine the iron content in soft drinks by colorimetry.
Recently Updated Pages
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Know The Difference Between Fluid And Liquid

Difference Between Crystalline and Amorphous Solid: Table & Examples

Types of Solutions in Chemistry: Explained Simply

Hess Law of Constant Heat Summation: Definition, Formula & Applications

Disproportionation Reaction: Definition, Example & JEE Guide

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

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

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

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2025-26

CBSE Notes Class 11 Chemistry Chapter 6 - Equilibrium - 2025-26

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

