Chapter 4, Problem 4.96QP

(a)

Interpretation Introduction

Interpretation:

Mass of the precipitate after the solutions are mixed has to be calculated.

Concept introduction:

Precipitation reaction:

When the two solutions mixed together and resultant compound settles down in the solution as a precipitate is called precipitation reaction.

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

$\text{Mass}\text{=}\text{Mole}\text{×}\text{Molar}\text{mass}$

(b)

Interpretation Introduction

Interpretation:

Concentration of the final solution has to be calculated.

Concept introduction:

Precipitation reaction:

When the two solutions mixed together and resultant compound settles down in the solution as a precipitate is called precipitation reaction.

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

$\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

Mass:

The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

$\text{Mass}\text{=}\text{Mole}\text{×}\text{Molar}\text{mass}$