Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare $250mLof0.100MNaBr$ solution has to be calculated.

Concept Introduction:

Molarity $\left(M\right)$ can be calculated using following equation,

    $M=\frac{molsolute\left(n\right)}{Lsolution\left(V_L\right)}$

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    $\text{number}\text{of}\text{mol,}\text{n}\text{=}\frac{\text{mass}\text{of}\text{solute}\left(\text{w}\right)}{\text{molecular}\text{mass}\left(\text{M}\right)}$

On combining the above two equations another expression for molarity is obtained as below.

    $Molarity=\frac{\text{mass}\text{of}\text{solute}}{molecularmass\times volume\left(L\right)}$

Answer to Problem 6.54QP

Number of grams of solute needed to prepare $250mLof0.100MNaBr$ solution is $2.57g.$

Explanation of Solution

Molecular mass of sodium bromide is $102.894g/mol.$

Number of grams of solute needed to prepare $250mL\left(0.250L\right)of0.100MNaBr$ solution can be calculated as follows,

    $\begin{array}{c}Molarity=\frac{\text{mass}\text{of}\text{solute}}{molecularmass\times volume\left(L\right)}\\ \\ \text{mass}\text{of}\text{solute}=\left(Molarity\right)\times \left(molecularmass\right)\times \left(volume\left(L\right)\right)\\ \\ =\left(0.100M\right)\times \left(102.894g/mol\right)\times \left(0.250L\right)\\ \\ =2.57g.\end{array}$

Amount solute required is $2.57g.$

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare $250mLof0.200MKOH$  solution has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.54QP

Number of grams of solute needed to prepare $250mLof0.200MKOH$ solution is $2.8g.$

Explanation of Solution

Molecular mass of $KOH$ is $56.1056g/mol.$

Number of grams of solute needed to prepare $250mLof0.200MKOH$ solution can be calculated as follows,

    $\begin{array}{c}Molarity=\frac{\text{mass}\text{of}\text{solute}}{molecularmass\times volume\left(L\right)}\\ \\ \text{mass}\text{of}\text{solute}=\left(Molarity\right)\times \left(molecularmass\right)\times \left(volume\left(L\right)\right)\\ \\ =\left(0.200M\right)\times \left(56.1056g/mol.\right)\times \left(0.250L\right)\\ \\ =2.8g.\end{array}$

Amount solute required is $2.8g.$