Chapter 13, Problem 13.16QP

(a)

Interpretation Introduction

Interpretation: The molality for $\text{2}\text{.50 M NaCl}$ solution has to be calculated.

Concept introduction

Molality: Molality is defined as number of moles of the solute present in the specified amount of the solvent in kilograms.

$\text{Molality =}\frac{\text{number of moles of the solute}}{\text{kg of solvent}}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Answer to Problem 13.16QP

Molality of $\text{2}\text{.50 M NaCl}$ solution is $\text{2}\text{.68m}$.

Explanation of Solution

Given data: Strength of $\text{NaCl}$ solution $\text{=}\text{2}\text{.50M}$

Density of $\text{NaCl}$ solution $\text{=}\text{1}\text{.08}\text{g /mL}$

Calculate mass of water:

$\text{mass of 1L solution =1000 }\cancel{\text{mL}}\text{×}\frac{\text{1}\text{.08g}}{\text{1}\cancel{\text{mL}}}\text{=}\text{1080g}$

$\begin{array}{c}\text{mass of water =1080g -}\left(\text{2}\text{.50 }\cancel{\text{mol}}\text{ NaCl×}\frac{\text{58}\text{.44g NaCl}}{\text{1}\cancel{\text{mol}}\text{ NaCl}}\right)\\ \text{=934}\text{g}\\ \text{=0}\text{.934}\text{kg}\end{array}$

The mass of solvent (water) determined by calculating the mass of solution then subtracting the solute mass with solution mass.

Calculate molality of the solution:

$\text{molality =}\frac{\text{2}\text{.50}\text{mol NaCl}}{\text{0}\text{.934}{\text{kg H}}_{\text{2}}\text{O}}\text{=}\text{2}\text{.68}\text{m}$

By plugging the value of $\text{NaCl}$ moles and amount of solvent to molality formula, the molality of the solution has been calculated.

(b)

Interpretation Introduction

Interpretation: The molality for $\text{48}\text{.2 mass percent of KBr}$ solution has to be calculated.

Concept introduction

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Percent by mass: It is the ratio of solute mass to the solution mass which finally multiplied by $\text{100}$. The formula for percent by mass calculation is as follows,

$\text{Percent mass =}\frac{\text{Mass}\text{of}\text{the}\text{element}}{\text{total}\text{mass}\text{of}\text{the}\text{compound}}\text{×100\%}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The $S.I.$ unit of mass is $kg$.

Answer to Problem 13.16QP

Molality of $\text{48}\text{.2 percent mass KBr }$ solution is $\text{7}\text{.82 m}$.

Explanation of Solution

Given data: Mass percent for $\text{KBr }$$=48.2$

From mass percent it is clear that $\text{100}\text{g}$ solution consists $48.2g$ of solute $\left(\text{KBr}\right)$ and $51.8g$ solvent $\left({\text{H}}_{\text{2}}\text{O}\right)$ since $\text{100 - 48}\text{.2 = 51}\text{.8}$.

Calculate moles of$\text{KBr}$:

Molecular mass of $\text{KBr }$$\text{=119}\text{.0}\text{g}$

$\text{KBr moles = 48}\text{.2 }\cancel{\text{g}}\text{ KBr ×}\frac{\text{1mol KBr}}{\text{119}\text{.0 }\cancel{\text{g}}\text{ KBr}}\text{= 0}\text{.405 mol KBr}$

By plugging the value of mass percent of the solution and molecular mass of solute, the moles of $\text{KBr }$ was calculated.

Calculate mass of water:

$\text{Mass of water = 51}\text{.8}\cancel{\text{g}}{\text{ H}}_{\text{2}}\text{O × }\frac{\text{1kg}}{\text{1000}\cancel{\text{g}}}\text{= 0}{\text{.0518 kg H}}_{\text{2}}\text{O}$

Calculate molality of the solution:

$\text{molality = }\frac{\text{0}\text{.405mol KBr}}{\text{0}{\text{.0518kg H}}_{\text{2}}\text{O}}\text{= 7}\text{.82m}$

By plugging the value of $\text{KBr }$ moles and amount of water into molality formula, the molality of the solution was calculated.