Q. 9.105 - An Introduction to General, Organic, and Biological Chemistry

Question

How many grams of solute are in each of the following solutions? 

Part a. $2.5 \mathrm{L}$ of a $3.0\mathrm{MAl}{\left({\mathrm{NO}}_3\right)}_3$ solution

Part b. $75 \mathrm{mL}$ of a $0.50{\mathrm{MC}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ solution

Part c. $235 \mathrm{mL}$of a $1.80 M \mathrm{LiCl}$ solution 

Step-by-Step Solution

Answer

Grams of solute the in each of the following solution:

a. $2.5 \mathrm{L}$of a  $3.0\mathrm{MAl}{\left({\mathrm{NO}}_3\right)}_3$solution $=1597.35 \mathrm{g}$

b. $75 \mathrm{mL}$ of a $0.50{\mathrm{MC}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ solution $\approx 6.8 \mathrm{g}$

c. $235 \mathrm{mL}$ of a $1.80 M \mathrm{LiCl}$ solution $\approx 17.9 g \mathrm{of} \mathrm{LiCl}$

Step 1: Introduction (Part a)

To calculate how many grams of solute in the given following solution:

Step 2: Given:

To calculate grams of solute in given 

a. $2.5 L$ of a $3.0\mathrm{MAl}{\left({\mathrm{NO}}_3\right)}_3$ solution 

b.$75 mL$ of a $0.50{\mathrm{MC}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ solution

c. $235 \mathrm{mL}$ of a $1.80 M \mathrm{LiCl}$solution 

Step 3: (part a) Calculation of Molarity:

Molarity, $\mathrm{M}=\frac{\text{ Molar of solute }}{\text{ Volume of solvent in Liters }}$

Moles of solute $=$ (Molarity , $M$) [volume of solvent in liters]

$=(3.0 \mathrm{mol}/\mathrm{L})(2.5 \mathrm{L})$

$=7.5mol$

Grams of solute$=(7.5 \mathrm{mol})($ Molar mass of solute $)$

$=(7.5 \mathrm{mol})(212.99 \mathrm{g}/\mathrm{mol})$

$=1597.35 \mathrm{g}$

Step 4: (part b) To calculate Moles of solute:

Moles of solute $=$ (Molarity, $M$) (Volume of solvent in liters)

$=(0.50 \mathrm{mol}/\mathrm{L})\left(75 \mathrm{mL}\left(\frac{1 \mathrm{L}}{1000 \mathrm{mL}}\right)\right)$

$=(0.50 \mathrm{mol}/\mathrm{L})(0.075 \mathrm{L})$

$=0.0375mol$

Mass of solute $=(0.0375 \mathrm{mol})($ Molar mass of solute$)$

$=(0.0375 \mathrm{mol})\left(\frac{180.0 \mathrm{g}}{1 \mathrm{mol}}\right)$                                                                             $=6.75 \mathrm{g}$

$\approx 6.8 \mathrm{g}$ of ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

Step 5: (part c) To calculate moles of solute :

Moles of solute $=($Molarity,$M)$$($ volume of  the solvent in liters$)$

$=(1.80\mathrm{M})\left(235 \mathrm{mL}\left(\frac{1 \mathrm{L}}{1000 \mathrm{mL}}\right)\right)$

$=(1.80\mathrm{M})(0.235 \mathrm{L})$

$=(1.80 \mathrm{mol}/\mathrm{L})(0.235 \mathrm{L})$

$=0.423 \mathrm{mol}$

Mass of solute $=($Moles of the solute$) ($ Molar mass the of solute $)$

$=(0.423 \mathrm{mol})(42.4162 \mathrm{g}/\mathrm{mol})$

$=17.9420 \mathrm{g}$

$\approx 17.9 \mathrm{g} \mathrm{of} \mathrm{LiCl}$