SOLUTIONS 1
1 (b)
( ) glucose =( ) unknown compound
0.05=
n= =2 (e.f.m. for )
so, molecular formula=
2 (a)
i=van,t Hoff factor
m= molality and
and = freezing point depression constant
For 0.01 molal NaCl solution
-----(i)
For 0.02 molal urea solution
-----(ii)
From Eqs (i) and (ii)
3 (b)
van’t Hoff factor i=[1+(y-1) ]
where y is the number of ions from one mole solute, (in this case =3), the degree of dissociation.
i=(1+2
4 (c)
6 (a)
7 (a)
mole-fraction
In 10 mole solubility is
8 (d)
Mixture solution boil at 1 atm = 760 mm = total pressure.
0.5, mol% of A =50%
9 (b)
Depression in freezing point is colligative property. The solute which produces highest number of ions will
have minimum freezing point .
1. One molal NaCl aqueous solution
2 ions/molecule
2. One molal solution
3 ions/molecule
3. One molalKCl aqueous solution
KCl
2 ions/molecule
4. One molal urea aqueous solution no dissociation
CaC solution has highest number of ions
It has lowest freezing point.
10 (a)
Ideal solution H =0
=0
= =
11 (c)
Given, vapour pressure of benzene,
=640 mm Hg
Vapour pressure of solution,
p=600 mm Hg
Weight of solute, w= 2.175 g
Weight of benzene, W= 39.08 g
Molecular weight of benzene,
M=78 g
Molecular weight of solute, m=?
According to Raoult’s law,
m=
m=69.60
12 (b)
13 (a)
Van’t Hoff’s factor (i)=4 {
Molality =
(As freezing point of water is )
14 (b)
molecular formula of the compound is
15 (b)
For
16 (c)
Number of moles = Molarity Volume (in L)
Number of moles of
= 10 moles
17 (c)
We know that 1 g equivalent weight of NaOH = 40 g
40 g of NaOH = 1 g eq. Of NaOH
0.275 g of NaOH = 0.275 eq.
=6.88 meq
(HCl) (NAOH)
( meq = NV)
18 (d)
Equivalent weight of
Oxidation number of Cr in
2[+1]+2(x)+7(-2)= 0
2+2x-14 = 0
2x=12
x=6
Equivalent weight =
= 4.903 g
20 (a)
According to Raoult’s law
21 (c)
Two solutions are isotonic if their osmotic pressure are equal.
( )
At a given temperature,
( )
Cane sugar unkown
=68.4 g
22 (d)
According to Raoult’s law,
or
23 (a)
One molar (1 M) aqueous solution is more concentrated than one molal aqueous solution of the same solute. In
solution, provides three ions. While NaCl provides two ions. Hence, vapour pressure of solution of
NaCl is higher (as it gives less ions). Therefore, 1 molal NaCl will have the maximum vapour pressure.
24 (b)
The number of moles or gram molecules of solute dissolved in 1000 g of solvent = molality
117 g NaCl = 2 mol
Hence, concentration of solution = 2 molal
25 (a)
In solution the KCl and produces same number of ions in solution.
Both produced two ions in solution.
So, ionic strength of a solution is combined ionic strength of both of the salt.
=0.1+.02=0.3 mol/kg
26 (b)
Ions at equilibrium 1-
Given,
Mass of HBr = 8.1 g
Mass of =100 g
( ) = degree of ionization = 90%
m(molality) =
=1+90/100
=1.9
= 1.9 1.86
= 3.53
(depression in freezing point) = freezing point of water - freezing point of solution 3.534 = 0 – freezing
point of solution.
Freezing point of solution = -3.53
27 (d)
28 (c)
29 (c)
For ideal solution,
31 (c)
Molarity =
moles of urea =
weight of solution =weight of solution + weight of solute
=1000 + 120 =1120 g
=0.974 K
Molarity = = 2.05 M
32 (a)
=CRT
Hence, C =0.2 M
R =0.082 L atm
T =27+273 = 300 K
=0.2
=4.92 atm.
33 (c)
34 (a)
Molarity = normality
Given, normality of solution =0.2 N
Equivalent weight = M
Molecular weight 2 M ( )
Molarity
35 (b)
Given =19.8 mm
According toRaoult’s law
or 198-10 p = 19.8 0.1
10 p = 198-1.98
10 p =196.02
p = 19.602 mm
36 (a)
37 (b)
i for HBr=1 +
where, =degree of dissociation
i=1+0.9=1.9
1.9 1.86
=3.534
Freezing point =-3.534
38 (d)
Moles =
Given, mass of = 50 g
molecular mass of = 342
Moles of mol
39 (b)
Let the volume of 0.4 M HCl is and that of 0.9 M HCl is
We know that,
(Mixture) (for 0.4 M HCl) ( for 0.9 M HCl)
[ 1m HCl = 1N HCl]
40 (b)
n-heptane and ethanol forms non-ideal solution. In pure ethanol, Molecules are hydrogen bonded. On adding n-
heptane, its molecules get in between the host molecules and break sme of the hydrogen bonds between them.
Due to weaking of interactions, the solution shows positive deviation from Raoult’s law.
