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Problem : One kg solution has 0.6 % of urea (w/w) in it. If the molar mass of urea is 60 $gm/mo{l}^{-1}$ , then determine the moles of urea present in the solution.
Solution : The mass of solute (oxalic acid) is obtained as :
$${g}_{B}=\frac{xX{g}_{S}}{100}=\frac{0.6X1000}{100}=60\phantom{\rule{1em}{0ex}}gm$$
$${n}_{B}=\frac{{g}_{A}}{{M}_{O}}=\frac{60}{60}=1$$
Sulphuric acid is formed by passing $S{O}_{3}$ gas through water in accordance with following chemical equation :
$$S{O}_{3}+{H}_{2}O\to {H}_{2}S{O}_{4}$$
When all water molecules combine to form sulphuric acid, there remains free $S{O}_{3}$ molecules. Oleum is the name given to this mixture of concentrated sulphric acid solution and free $S{O}_{3}$ . From the point of view reaction between oleum and other solution, $S{O}_{3}$ molecules are as good as sulphuric acid molecule as it reacts with available water molecules to form sulphuric acid molecule.
If an oleum solution has x percent of free $S{O}_{3}$ (x gm in 100 gm of oleum), then the equivalent amount of sulphuric acid is calculated using mole concept :
$$\Rightarrow \text{1 mole of}\phantom{\rule{1em}{0ex}}S{O}_{3}\equiv \text{1 mole of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$$
$$\Rightarrow \text{80 gm of}\phantom{\rule{1em}{0ex}}S{O}_{3}\equiv \text{98 gm of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$$
${H}_{2}S{O}_{4}$ that can be formed from x gm of SO3 is $\frac{98x}{80}$ . Total mass of ${H}_{2}S{O}_{4}$ is sum of ${H}_{2}S{O}_{4}$ in the liquid form and ${H}_{2}S{O}_{4}$ that can be formed when ${H}_{2}S{O}_{4}$ reacts with water :
$$\Rightarrow \text{total mass of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}=100-x+\frac{98x}{80}=100+\frac{18x}{80}$$
Considering equivalent ${H}_{2}S{O}_{4}$ mass in calculation, the mass percentage of ${H}_{2}S{O}_{4}$ in oleum is given as :
$$\Rightarrow y=\frac{\text{Total mass of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}}{\text{mass of oleum}}=\frac{100+\frac{18x}{80}}{100}X100=100+\frac{18x}{80}$$
Clearly, equivalent ${H}_{2}S{O}_{4}$ mass percentage is more than 100 %.
The volume percentage is expressed as :
$$\text{Volume percentage}\left(x\right)=\frac{\text{Volume of solute}\left(B\right)}{\text{Volume of solution}\left(A+B\right)}X100$$
$$\Rightarrow \text{Volume percentage}\left(x\right)=\frac{{V}_{B}}{{V}_{S}}X100=\frac{{V}_{B}}{{V}_{A}+{V}_{B}}X100$$
The strength of solution is expressed as :
$$\text{Strength of solution}\left(S\right)=\frac{\text{Mass of solute (B) in grams}}{\text{Volume of solution in litres}}$$
$$\Rightarrow S=\frac{{g}_{B}}{{V}_{L}}$$
The unit of strength is “grams/liters”. Note that strength of solution is not a percentage – rather a number The symbol ${V}_{L}$ denotes volume of solution in litres, whereas ${V}_{\mathrm{CC}}$ denotes volume of solution in cc. If volume is expressed in cc, then the formula of strength is :
$$\Rightarrow S=\frac{{g}_{B}X1000}{{V}_{cc}}$$
Problem : Determine the numbers of moles of sulphuric acid present in 500 cc of 392 gm/litre acid solution.
Solution : In order to find the moles of sulphuric acid, we need to find its mass in the given volume.
$${g}_{B}=SX{V}_{L}=392X0.5=196\phantom{\rule{1em}{0ex}}gm$$
The moles of ${H}_{2}S{O}_{4}\phantom{\rule{1em}{0ex}}\left({M}_{0}=2+32+4X16=98\right)$ is :
$${n}_{B}=\frac{196}{98}=2$$
The conversion of concentration of the solution into molar mass of the solute is a two steps process. In the first step, we calculate the mass of the solute and then in second step we divide the mass of the solute by molecular weight to determine the moles of solute present in the solution. We can think of yet a direct measurement of molar concentration. This will enable us to calculate moles of solute in one step.
Mole fraction of solute ( ${\chi}_{B}$ ) is defined as :
$$\text{Molefraction}\left({\chi}_{B}\right)=\frac{\text{Mole of solute}\left(B\right)}{\text{Mole of solvent}\left(A\right)+\text{Mole of solute}\left(B\right)}$$
$$\Rightarrow {\chi}_{B}=\frac{{n}_{B}}{{n}_{A}+{n}_{B}}$$
Similarly, mole fraction of solvent (B) or other component of solution is :
$$\Rightarrow {\chi}_{A}=\frac{{n}_{A}}{{n}_{A}+{n}_{B}}$$
Clearly,
$$\Rightarrow {\chi}_{A}+{\chi}_{B}=1$$
Problem : The mass fraction of ethyl alcohol in a sample of 1 kg of aqueous ethyl alcohol solution is 0.23. Determine mole fraction of ethyl alcohol and water in the solution.
Solution : The mass of ethyl alcohol and water are calculated as :
$$\Rightarrow {g}_{B}=\text{mass fraction}X\text{mass of solution}=0.23X1000=230\phantom{\rule{1em}{0ex}}gm$$
$$\Rightarrow \text{mass of solvent}={g}_{A}=\text{mass of solution}-\text{mass of ethyl alcohol}=1000-230=770\phantom{\rule{1em}{0ex}}gm$$
The moles of of ethyl alcohol and water are :
$$\Rightarrow {n}_{B}=\frac{230}{{M}_{{C}_{2}{H}_{5}OH}}=\frac{230}{2X12+5X1+16+1}=\frac{230}{46}=5$$
$$\Rightarrow {n}_{A}=\frac{770}{{M}_{{H}_{2}O}}=\frac{230}{18}=\frac{770}{18}=42.8$$
The mole fraction of ethyl alcohol and water are :
$$\Rightarrow {\chi}_{B}=\frac{{n}_{B}}{{n}_{A}+{n}_{B}}=\frac{5}{42.8+5}=\frac{5}{47.8}=0.104$$
$$\Rightarrow {\chi}_{A}=1-0.104=0.896$$
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