Chemistry: Solutions - Evaluate Yourself - with Solution

**Evaluate Yourself**

1)
If 5.6 g of KOH is present in (a) 500 mL and (b) 1 litre of solution, calculate
the molarity of each of these solutions.

**Solution:**

mass
of KOH = 5.6 g

no.
of moles = 5.6/56 = 0.1 mol

molarity
= number of moles of solute / volume of solution (in L)

i)
Volume of the solution = 500 ml = 0.5 L

molarity
= number of moles of solute / volume of solution (in L)

Molarity
= 0.1/0.5 = 0.2M

ii)
Volume of the solution = 1 L

molarity
= number of moles of solute / volume of solution (in L)

Molarity
= 0.1/1 = 0.1M

2)
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of
glucose and water.

**Solution:**

mass
of glucose = 2.82 g

no.
of moles of glucose = 2.82/180=0.016

mass
of water = 30 g

=30/18
= 1.67

X_{H}_{2}_{0 }= 1.67 /
[1.67+0.016] = 1.67 / 1.686 = 0.99

X_{H}_{2}_{0 }+ X_{glucose } =1

0.99
+
X_{glucose } =1

X_{glucose } = 1 - 0.99

X_{glucose }= 0.01

3)
The antiseptic solution of iodopovidone for the use of external application
contains 10 % w/v of iodopovidone. Calculate the amount of iodopovidone present
in a typical dose of 1.5 mL.

**Solution:**

10%
W/V means that 10g of solute in 100ml solution.

amount
of iodopovidone in 1.5 ml = 10g/100ml x 1.5ml = 0.15g

4)
A litre of sea water weighing about 1.05 kg contains 5 mg of dissolved oxygen
(O_{2}). Express the concentration of dissolved oxygen in ppm.

**Solution:**

[ mass of dissolved solid / mass of water ] x
10^{6}

5x10^{-3}/1.05x10^{3} x 10^{6}
= 4.76ppm

5)
Describe how would you prepare the following solution from pure solute and
solvent

(a)
1 L of aqueous solution of 1.5 M CoCl_{2}.

(b)
500 mL of 6.0 % (V/V) aqueous methanol solution.

**Solution:**

(a)
mass of 1.5 moles of CoCl_{2} = 1.5 × 129.9

=
194.85 g

194.85
g anhydrons cobalt chloride is dissolved in water and the solution is make up
to one litre in a standard flask.

b.
6% V/Vaqeous solution contains 6g of methanol in 100 ml solution.

∴ To prepare 500 ml of 6% V/V solution of methanol 30g
methanol is taken in a 500 ml standard flask and required quantity of water is
added to make up the solution to 500 ml.

6)
How much volume of 6 M solution of NaOH is required to prepare 500 mL of 0.250
M NaOH solution.

**Solution:**

C_{1}V_{1}
= C_{2}V_{2}

6
M (V_{1}) = 0.25 M × 500 ml

V_{1}
= 0.25x500/6

V_{1}
= 20.83 mL

7)
Calculate the proportion of O_{2} and N_{2} dissolved in water
at 298 K. When air containing 20% O_{2} and 80% N_{2} by volume
is in equilibrium with it at 1 atm pressure. Henry’s law constants for two
gases are K_{H}(O_{2}) = 4.6 x 10^{4} atm and K_{H}
(N_{2}) = 8.5 x 10^{4} atm.

**Solution:**

8)
Explain why the aquatic species are more comfortable in cold water during
winter season rather than warm water during the summer.

9)
Calculate the mole fractions of benzene and naphthalene in the vapour phase
when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39
g of benzene. It is given that the vapour pressure of pure benzene is 50.71
mmHg and the vapour pressure of pure naphthalene is 32.06 mmHg at 300 K.

**Solution:**

PmmHgPmmHg

Pº_{Pure
benzene} = 50.71 mm Hg

Pº_{naphthalene}
= 32.06 mm Hg

Number
of moles of benzene = 39/78 = 0.5 mol

Number
of moles of napthalene = 128/128 = 1 mol

mole
fraction of benzene = 0.5/1.5 = 0.33

mole
fraction of napthalene = 1 – 0.33 = 0.67

Partial
vapour pressure of benzene = Pº_{benzene} × mole fraction of benzene

=
50.71 × 0.33

=
16.73 mm Hg

Partial
vapour pressure of napthalene = 32.06 × 0.67

=
21.48 mm Hg

Mole
fraction of benzene in vapour phase =

=
16.73/[16.73+21.48] = 16.73/38.21 = 0.44

Mole
fraction of napthalene in vapour phase = 1 – 0.44 = 0.56

10)
Vapour pressure of a pure liquid A is 10.0 torr at 27°C . The vapour pressure
is lowered to 9.0 torr on dissolving one gram of B in 20 g of A. If the molar
mass of A is 200 then calculate the molar mass of B.

**Solution:**

P_{A}º
= 10 torr , P_{solution} = 9 torr

W_{A}
= 20 g

W_{B}
= 1 g

M_{A}
= 200 g mol^{–1}

M_{B}
= ?

∆P/
P_{A}º = W_{B}xM_{A} / M_{B}xW_{A}

[10-9]/10
= [1x200]/M_{B}x200

M_{B}=
200/20 x 10 = 100g mol^{-1}_{}

11)
2.56 g of Sulphur is dissolved in 100g of carbon disulphide. The solution boils
at 319. 692 K . What is the molecular formula of Sulphur in solution The
boiling point of CS_{2} is 319. 450K. Given that K_{b} for CS_{2}
= 2.42 K Kg mol^{-1}

**Solution:**

W_{2}
= 2.56 g

W_{1}
= 100 g

T
= 319.692 K

K_{b}
= 2.42 K Kg mol^{–1}

∆T_{b}
= (319.692 – 319.450) K

=
0.242 K

M_{2}
= K_{b}xW_{2}x100 / ∆T_{b}xW_{1}

=
2.42x2.56x1000 / 0.242x100

M_{2}
= 256g mol^{-1}

Molecular
mass of sulphur in solution = 256 g mol^{–1}

atomic
mass of one mole of sulphur atom = 32

No.
of atoms in a molecule of sulphur =256/32=8

Hence
molecular formula of sulphur is S_{8}.

12)
2g of a non electrolyte solute dissolved in 75 g of benzene lowered the
freezing point of benzene by 0.20 K. The freezing point depression constant of
benzene is 5.12 K Kg mol^{-1}. Find the molar mass of the solute.

**Solution:**

W_{2}
= 2g

W_{1}
= 75 g

ΔT_{f}
= 0.2 K

K_{f}
= 5.12 K Kg mol^{–1}

M_{2}
= ?

M_{2}=
682.66 mol^{-1}_{}

13)
What is the mass of glucose (C_{6}H_{12}O_{6}) in it
one litre solution which is isotonic with 6 g L^{-1} of urea (NH_{2}CONH_{2})
?

**Solution:**

Osmotic
pressure of urea solution (π_{1})=CRT

14)
0.2 m aqueous solution of KCl freezes at -0.68ºC calculate van’t Hoff
factor. k_{f} for water is 1.86
K kg mol^{-1}.

**Solution:**

i
= observed property / Theoritical property (calculated)

Given
ΔT_{f} = 0.680 K

m
= 0.2 m

ΔT_{f}
(observed) = 0.680 K

ΔT_{f}
(calculated) = K_{f} . m

=
1.86 K Kg mol^{–1} × 0.2 mol Kg^{–1}

=
0.372 K

i = (∆T_{f}) observed / (∆T_{f}) calculated = 0.680 K / 0.372 K = 1.82

Tags : 11th Chemistry : Solutions

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail