Normality , Molarity , Molality source : solution- definition. molarity.unit conversion .flow chart and ….

## 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

## 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 200 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.