# 0.2 Molar concentration  (Page 2/3)

 Page 2 / 3

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 :

${g}_{B}=\frac{xX{g}_{S}}{100}=\frac{0.6X1000}{100}=60\phantom{\rule{1em}{0ex}}gm$

${n}_{B}=\frac{{g}_{A}}{{M}_{O}}=\frac{60}{60}=1$

## Mass percentage of oleum

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 :

$⇒\text{1 mole of}\phantom{\rule{1em}{0ex}}S{O}_{3}\equiv \text{1 mole of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

$⇒\text{80 gm of}\phantom{\rule{1em}{0ex}}S{O}_{3}\equiv \text{98 gm of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}$

${H}_{2}S{O}_{4}$ 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 :

$⇒\text{total mass of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}=100-x+\frac{98x}{80}=100+\frac{18x}{80}$

Considering equivalent ${H}_{2}S{O}_{4}$ mass in calculation, the mass percentage of ${H}_{2}S{O}_{4}$ in oleum is given as :

$⇒y=\frac{\text{Total mass of}\phantom{\rule{1em}{0ex}}{H}_{2}S{O}_{4}}{\text{mass of oleum}}=\frac{100+\frac{18x}{80}}{100}X100=100+\frac{18x}{80}$

Clearly, equivalent ${H}_{2}S{O}_{4}$ mass percentage is more than 100 %.

## Volume percentage (v/v)

The volume percentage is expressed as :

$\text{Volume percentage}\left(x\right)=\frac{\text{Volume of solute}\left(B\right)}{\text{Volume of solution}\left(A+B\right)}X100$

$⇒\text{Volume percentage}\left(x\right)=\frac{{V}_{B}}{{V}_{S}}X100=\frac{{V}_{B}}{{V}_{A}+{V}_{B}}X100$

## Strength of solution (w/v)

The strength of solution is expressed as :

$\text{Strength of solution}\left(S\right)=\frac{\text{Mass of solute (B) in grams}}{\text{Volume of solution in litres}}$

$⇒S=\frac{{g}_{B}}{{V}_{L}}$

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 :

$⇒S=\frac{{g}_{B}X1000}{{V}_{cc}}$

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 :

${n}_{B}=\frac{196}{98}=2$

## Molar concentration

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

Mole fraction of solute ( ${\chi }_{B}$ ) is defined as :

$\text{Molefraction}\left({\chi }_{B}\right)=\frac{\text{Mole of solute}\left(B\right)}{\text{Mole of solvent}\left(A\right)+\text{Mole of solute}\left(B\right)}$

$⇒{\chi }_{B}=\frac{{n}_{B}}{{n}_{A}+{n}_{B}}$

Similarly, mole fraction of solvent (B) or other component of solution is :

$⇒{\chi }_{A}=\frac{{n}_{A}}{{n}_{A}+{n}_{B}}$

Clearly,

$⇒{\chi }_{A}+{\chi }_{B}=1$

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 :

$⇒{g}_{B}=\text{mass fraction}X\text{mass of solution}=0.23X1000=230\phantom{\rule{1em}{0ex}}gm$

$⇒\text{mass of solvent}={g}_{A}=\text{mass of solution}-\text{mass of ethyl alcohol}=1000-230=770\phantom{\rule{1em}{0ex}}gm$

The moles of of ethyl alcohol and water are :

$⇒{n}_{B}=\frac{230}{{M}_{{C}_{2}{H}_{5}OH}}=\frac{230}{2X12+5X1+16+1}=\frac{230}{46}=5$

$⇒{n}_{A}=\frac{770}{{M}_{{H}_{2}O}}=\frac{230}{18}=\frac{770}{18}=42.8$

The mole fraction of ethyl alcohol and water are :

$⇒{\chi }_{B}=\frac{{n}_{B}}{{n}_{A}+{n}_{B}}=\frac{5}{42.8+5}=\frac{5}{47.8}=0.104$

$⇒{\chi }_{A}=1-0.104=0.896$

anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!