Normality , Molarity , Molality

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**Solved problems of Normality , Molarity , Molality-**

**Q. 1 ) 49 grams of H2SO4 are present in 100 ml aqueous solution. What is the molarity of H2SO4 ?**

**Solution : w=49 gm. ,V = 100 ml. =0.1 liter , m of H2SO4 =2 +32+16 x 4=98**

** M =w/ mV(l)**

=49/(98 x 0.1)

*M= 5 mole / liter*

## Q.2 ) An aqueous solution of HCl is 38% by mass & its density is 1.19 gm/ml. calculate the molality & molarity of solution.

Solution :

Mass of solute ‘w’ = 38 gram

Mass of Solution= 100 gram

Mass of Solvent ‘W’= 100-38

= 62 gm = 0.062 kg

m of HCl =1 +35.5 =36.5

molality = w/ mW(Kg.)

=38/(36.5 x 0.62)

**molality = 16.79 mole /Kg.**

Volume =mass /density=100/1.19 =84 ml. =0.084 liter

**M = w/ mV(l)**

=38 /(36.5 x 0.084)

** Molarity = 12.39 mole/liter**

## Q.3) 8 grams of solute are dissolved in 10 liter of the solution. Calculate the concentration in ppm.

## Solution :

w =8 gm. =8 x 1000 =8000 mg.,V =10 liter

concentration in ppm =weight of solute (mg.)/volume of solution (l)

8000/10 =800 ppm

**concentration = 800 ppm**

## Q.4 ) A mixture of C2H5OH & water contains 54% water by weight. Calculate the mole fraction of alcohol & Water in this mixture ?

Solution :

w of H2O = 54 gm.

w of C2H5OH = 100-54 = 46 gm.

no. of moles of water = 54/18

=3

no. of moles of alcohol = 46/46 = 1

total moles =3 +1 = 4

Xw = nw /total moles =3/4=0.75

**mole fraction of water =0.75**

X ethyl alcohol =n ethyl alcohol/total moles =1/4=0.25

**mole fraction of ethyl alcohol =0.25**

**Q.5) The density of a 2.03 M solution of acetic acid (molecular weight = 60) in water is 1.017 gm/ml. Calculate molality of the solution.**

** Solution:**

S = Molecular weight × Molarity

= 60 x 2.03

= 121.80 gm/l

Mass of solute (w) = 121.80 gm

v = 1 liter = 1000 ml

mass of solution = V×d = 1000×1.017

= 1017 gm

mass or weight of solvent =1017-121.80=895.2 gm. =0.8952 Kg.

**Molality ***=* w/mW(Kg.)

= 121.8 /(60 x 0.8952)

**Molality =2.27 mole/Kg.**

**Q.6) 4 gms of NaOH is dissolved in 200 ml of aqueous Solution. What will be the molarity of this solution ?**

**Solution :**

w =4 gms. , V=200 ml = 0.2 L , m = 23 +16 +1 = 40

**Molarity =w /mV (L)**

=4 /(40 x 0.2)

M** =0.5 mole /liter **

**Q.7) Calculate the quantity of Na2CO3 required to prepare 250 ml , M/20 solution.**

**Solution :**

w=? , m= 23 x 2+12 +16 x 3= 106 ,V =250 ml =0.25 L

**Molarity =w /mV (L)**

1/20 =w/ 106 x 0.25

** w =1.325 gm.**

**Q.8) A solution contains 410.3 grams of H2SO4 per liter of Solution at 20 ^{0} C. If the density is 1.243 gm/ml. What will be its Molarity & Molality.**

**Ans.**

w=410.3 gm.

V= 1 L

m = 2 x 1+32 +16 x 4= 98

Molarity = w/m V (l)

=410.3/98 x 1

**Molarity =4.187 mole/liter**

Mass of Solution = Volume of Solution × Density

= 1000 ×1.243

= 1243 gm.

weight of solvent = 1243 -410.3 = 832.7 gm = 0.8327 Kg.

**Molality ****= w/mW(Kg.)**

=410.3/98 x 0.8327

**Molality****= 5.028 mole/Kg.**

## Q. Calculate molarity of pure water. (density =1 gm/ml)

Solution =

Let,volume be 1000 ml

Mass of H2O = 1000×1=1000 gm.

V = 1000 ml. =1 liter

m =2 x 1 +16 =18

Molarity =w/m V(L)

=1000/18 x 1

**Molarity=55.56 mole/Kg.**