Chapter 1, Problem 1.AE

(a)

Interpretation Introduction

Interpretation:

Molarity of $500mL$ methanol solution of $25.00mL$ methanol in chloroform has to be calculated.

Concept Introduction:

Molarity is one of the parameters used to express concentration of a solution.  It is expressed as,

$\text{Molarity = }\frac{number\text{ of }molesof\text{ solute}}{volume\text{ of solution in L}}$

Answer to Problem 1.AE

Molarity of $500mL$ methanol solution made of $25.00mL$ methanol in chloroform is calculated as $1.235M.$

Explanation of Solution

Calculate the number of moles of solute methanol to determine.

Given that volume of methanol is $25.00\text{ mL}$ this can be converted to mass as,

$density\text{ = }\frac{mass}{volume}$

Rewriting the above equation to calculate mass of methanol,

$mass=volume\times density$

Given that density of methanol is $0.7914\text{ g/mL}\text{.}$ Therefore,

$\begin{array}{l}mass\text{ of methanol}\text{= 25}\text{.00 mL }\times 0.7914\text{ g/mL}\\ \text{= 19}\text{.785 g}\end{array}$

$\begin{array}{l}\text{no}\text{.of moles of methanol}=\frac{mass}{molar\text{ mass}}\\ =\frac{\text{19}\text{.785 g}}{32.04\text{ g/mol}}\\ =\text{ 0}\text{.6175 mol}\end{array}$

Volume of solution is $500\text{ mL(= 0}\text{.5 L)}$

Now calculate the molarity of methanol in $500mL$ solution containing $25.00\text{ mL (=25}\text{.00 g)}$ of methanol as shown below,

$\begin{array}{l}\text{Molarity }\text{= }\frac{number\text{ of }molesofmethanol}{volume\text{ of solution in L}}\\ =\frac{0.6175\text{ mol}}{0.5\text{ L}}\text{ = 1}\text{.235 M}\end{array}$

Conclusion

Conclusion: 

Given the density of methanol solution, its molarity is calculated using relevant formula.

(b)

Interpretation Introduction

Interpretation:

Molality of $500mL$ methanol solution of $25.00mL$ methanol in chloroform has to be calculated.

Concept Introduction:

Molality is one of the many parameters that is used to express concentration of a solution.  It is expressed as,

$\text{Molality = }\frac{number\text{ of }molesof\text{ solute}}{mass\text{ of solvent in kg}}$

Answer to Problem 1.AE

Molality of $500mL$ methanol solution made of $25.00mL$ methanol in chloroform is calculated as $0.873m.$

Explanation of Solution

Volume of solution is $500\text{ mL}\text{.}$  It is converted to mass –

$density\text{ = }\frac{mass}{volume}$

Rewriting the above equation to calculate mass of solution,

$mass=volume\times density$

Given that density of solution is $1.454\text{ g/mL}\text{.}$ Therefore,

$\begin{array}{l}mass\text{ of solution}\text{= 500 mL }\times 1.454\text{ g/mL}\\ \text{= 727 g}\end{array}$

$727\text{ g}$ of solution contains $25.00mL$ of methanol which is equivalent to $19.785\text{ g}$ methanol as calculated in previous step.  Therefore mass of solvent chloroform is,

$\begin{array}{l}\text{727 g of solution }\text{= 19}\text{.78 g of methanol + mass of chloroform}\\ \text{mass of chloroform = 727 g - 19}\text{.785 g }\\ \text{= 707}\text{.215 g = 0}\text{.707215 kg}\end{array}$

Number of moles of methanol as calculated in previous step is $0.6175\text{ mol}\text{.}$

Molality of methanol is calculated as,

$\begin{array}{l}\text{Molality of methanol}\text{= }\frac{number\text{ of }molesof\text{ methanol}}{mass\text{ of chloroform in kg}}\\ =\frac{\text{0}\text{.6175 mol}}{\text{0}\text{.707215}}\\ =\text{ 0}\text{.873 m}\end{array}$

Conclusion

Conclusion: 

Given the density of methanol solution, its molality is calculated using relevant formula.