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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.

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