Chapter 4, Problem 66QP

Determine how many grams of each of the following solutes would be needed to make $\text{2}\text{.50 }\times {\text{ 10}}^{\text{2}}\text{mL of a 0}\text{.100-Msolution: }\left(\text{a}\right)\text{ cesium iodide }\left(\text{CsI}\right)\text{, }\left(\text{b}\right)\text{ sulfuric acid }\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{, }\left(\text{c}\right)\text{ sodium carbonate }\left({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\text{, }\left(\text{d}\right)\text{ potassium dichromate }\left({\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\right)\text{, }\left(\text{e}\right)\text{ potassium permanganate }\left({\text{KMnO}}_{\text{4}}\right)\text{.}$

Interpretation Introduction

Interpretation:

The amount of solute needed to prepare a solution is to be determined.

Concept introduction:

The mass of a solute is the product of the number of moles and the molar mass of the solute.

$\text{Mass of solute}\left(\text{g}\right)\text{ = number of moles of the solute × molar mass of solute}\left(\text{g/mol}\right)$

The moles of a solute can be calculated by using the following expression:

$\text{Moles of solute}=\text{Molarity}\left(M\right)\times \text{volume}\left(\text{L}\right)$

Answer to Problem 66QP

Solution:

(a) $6.50\text{ g}$

(b) $\text{2}\text{.45 g}$

(c) $2.65\text{ g}$

(d) $7.35\text{ g}$

(e) $\text{3}\text{.95 g}$

Explanation of Solution

Given information:

The concentration of the solution is $0.10\text{0 M}$ and the volume of the solution is $2.50\times {10}^2\text{ mL}$ or $0.250\text{ L}$.

a) Cesium iodide $\text{CsI}$

Determine the number of moles of cesium iodide using the following expression:

$\text{Moles of solute}=\text{Molarity}\left(M\right)\times \text{volume}\left(\text{L}\right)$

Thus,

$\begin{array}{c}\text{Moles of CsI = 0}\text{.100 M × 0}\text{.250 L}\\ \text{= 0}\text{.0250 mol}\end{array}$

Now, calculate the mass of $\text{CsI}$ using the following expression:

$\text{Mass of solute}\left(\text{g}\right)\text{ = mol solute × molar mass of solute}\left(\text{g/mol}\right)$

Thus,

$\begin{array}{c}\text{Mass of CsI}\left(\text{g}\right)\text{ = 0}\text{.0250 mol × 259}\text{.809 g/mol CsI }\\ \text{= 6}\text{.50 g}\end{array}$

Hence, the mass of $\text{CsI}$ required to prepare a solution of the given concentration is $6.50\text{ g}$.

b) Sulfuric acid ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Determine the number of moles of sulfuric acid using the following expression: $\text{Moles of solute}=\text{Molarity}\left(M\right)\times \text{volume}\left(\text{L}\right)$

Thus,

$\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_4\text{ = 0}\text{.100M × 0}\text{.250 L}\\ \text{= 0}\text{.0250 mol}\end{array}$

Now, calculate the mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ using the following expression:

$\text{Mass of solute}\left(\text{g}\right)\text{ = mol solute × molar mass of solute}\left(\text{g/mol}\right)$

Thus,

$\begin{array}{c}{\text{Mass of H}}_2{\text{SO}}_4\left(\text{g}\right)\text{ = 0}\text{.0250 mol × 98}{\text{.079 g/mol H}}_2{\text{SO}}_4\\ \text{= 2}\text{.45 g}\end{array}$

Hence, the mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ required to prepare a solution of the given concentration is $\text{2}\text{.45 g}$.

c) Sodium carbonate ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

Determine the number of moles of sodium carbonate using the following expression:

$\text{Moles of solute}=\text{Molarity}\left(M\right)\times \text{volume}\left(\text{L}\right)$

Thus,

$\begin{array}{c}{\text{Moles of Na}}_2{\text{CO}}_3\text{ = 0}\text{.100M × 0}\text{.250 L}\\ \text{= 0}\text{.0250 mol}\end{array}$

Now, calculate the mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ using the following expression:

$\text{Mass of solute}\left(\text{g}\right)\text{ = mol solute × molar mass of solute}\left(\text{g/mol}\right)$

Thus,

$\begin{array}{c}{\text{Mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\left(\text{g}\right)\text{ = 0}\text{.0250 mol × 105}{\text{.99 g/mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\\ \text{= 2}\text{.65 g}\end{array}$

Hence, the mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ required to prepare a solution of the given concentration is $2.65\text{ g}$.

d) Potassium dichromate ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$

Determine the number of moles of potassium dichromate using the following expression:

$\text{Moles of solute}=\text{Molarity}\left(M\right)\times \text{volume}\left(\text{L}\right)$

Thus,

$\begin{array}{c}{\text{Moles of K}}_2{\text{Cr}}_2{\text{O}}_7\text{ = 0}\text{.100M × 0}\text{.250 L}\\ \text{= 0}\text{.0250 mol}\end{array}$

Now, calculate the mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ using the following expression:

$\text{Mass of solute}\left(\text{g}\right)\text{ = mol solute × molar mass of solute}\left(\text{g/mol}\right)$

Thus,

$\begin{array}{c}{\text{Mass of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\left(\text{g}\right)\text{ = 0}\text{.0250 mol × 294}{\text{.185 g/mol K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\\ \text{= 7}\text{.35 g}\end{array}$

Hence, the mass of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ required to prepare a solution of the given concentration is $7.35\text{ g}$.

e) Potassium permanganate ${\text{KMnO}}_{\text{4}}$

Determine the number of moles of potassium permanganate using the following expression:

$\text{moles of solute}=\text{Molarity}\left(M\right)\times \text{volume}\left(\text{L}\right)$

Thus,

$\begin{array}{c}{\text{Moles of KMnO}}_4\text{ = 0}\text{.100M × 0}\text{.250 L}\\ \text{= 0}\text{.0250 mol}\end{array}$

Now, calculate the mass of ${\text{KMnO}}_{\text{4}}$ using the following expression:

$\text{Mass of solute}\left(\text{g}\right)\text{ = mol solute × molar mass of solute}\left(\text{g/mol}\right)$

Thus,

$\begin{array}{c}{\text{Mass of KMnO}}_4\left(\text{g}\right)\text{ = 0}\text{.0250 mol × 158}{\text{.034 g/mol KMnO}}_4\\ \text{= 3}\text{.95 g}\end{array}$

Hence, the mass of ${\text{KMnO}}_{\text{4}}$ required to prepare a solution of the given concentration is $\text{3}\text{.95 g}$.