Here let me first introduce to the Naming system
Here subscript
A -> Stands for Solvent
B -> Solute
S -> Solution
$W_A$ = Mass of solvent
$M_A$ = Molar mass of solvent
$W_B$ = Mass of solute
$M_B $= Molar mass of solute
$n_A$ = no. of moles of solvent
$n_B$ = no. of moles of solute
$n_A = \frac {W_A}{M_A}$
$n_B = \frac {W_B}{M_B}$
$W_S$ = Mass of solution
$W_S = W_A + W_B$
$d_S$= density of solution
$d_S =\frac { W_S}{V_S}$
Question 1
6 gm of urea ($NH_2CONH_2$) is present in 250 ml of solution. Find molarity if the density of the solution is 1.1 gm/ml find molality.
Solution
$M_B$ = 60 gm
$W_B$ = 6 gm
$\text {Molarity} = \frac {W_B}{M_B} \times \frac {1000}{V_S} = .4 \ M $
Now
$d_S = 1.1 = \frac {W_S}{V_S}$
$W_S=275 $ g
Now
$W_S = W_A + W_B$
or $W_A= 269 $ g
Now
$\text {Molality} = \frac {W_B}{M_B} \times \frac {1000}{W_A} = \frac {(6 \times 1000)}{(60 \times 269)}= .371 m$
Question 2
10 gm of Ethylene glycol is dissolved in 500 gm of water. Find Molality. if the density of the solution is 1.2 gm/ml then find molarity.
Solution
for $H_2O$
$W_A$ = 500 gm
For Ethylene glycol($(CH_2OH)_2$)
$W_B$ = 10 gm,$M_B$ = 62 gm
Now $\text {Molality} = \frac {W_B}{M_B} \times \frac {1000}{W_A} = .322$ m
Now for $V_S$ we have
$d = W_S/V_S = (W_A+ W_B)/V_S $
$1.2 = (W_A+ W_B)/V_S $
$V_S = 510/1.2 =425$ ml
Now
$\text {Molarity} = \frac {W_B}{M_B} \times \frac {1000}{V_S} = .379 \ M $
Question 3
Find the molarity of 5 m aqueous glucose solution of density 1.1 gm/ ml
Solution
Glucose($C_6H_{12}O_6$)
$M_B$ = 180 gm ,Molality = 5m
$d_S = 1.1 $gm/ml
Now
$m = n_B/W_A $
$m = (W_A \times 1000)/(M_B \times W_A )$
When $W_A$ = 1000 gm then
$n_B = 5$
So,
$W_B/M_B =5$
$W_B$ = 900 gm
Also,
$d_S = W_S/V_S $
So, $V_S = (W_A+ W_B)/d_S = 1900/1.1$ ml
Now, Molarity = $(W_B \times 1000)/(M_B \times V_S )$
$=(900 \times 1000 \times 1.1)/(180 \times 1900)=2.89$M
Question 3
Find the molarity of a sample of 68% HNO
_{3} by mass in aqueous solution. If density of the solution in 1.504 gm/ml.
Solution
For HNO
_{3}
$M_B$ = 63 gm
$68% = W_B/W_S =68/100$
$W_B$ = 68 gm
Also,
$d_S = W_S/V_S $
$1.504 = 100/V_S $
$V_S = 100/1.504$
$M = (W_B \times 1000)/(M_B \times V_S )= 16.23 $M
Question 4
A sample of water contains 2 mg of a harmful substance in 1,000 kg of drinking water. Express the conc. of solution in ppm.
Solution
$ppm = W_B/W_S \times 10^6$
$= (2 \times 10^6 \times 10^{-3})/(1000 \times 1000 ) = 2 \times 10^{-3}$ ppm.
Question 5
A sample of drinking water is contaminated with chloroform. The label of contamination is 15 ppm. Express this conc. in terms of:
(i)% by mass of chloroform
(ii)Molality
Question 6
A solution of glucose in water is labeled as 10% (W/W). Find the molality & mole fraction. If the density of the solution is 1.2 gm/ml then find molarity.
Question 7
500ml of water is added to 2l of 1M. Find the molarity of the solution after dilution.
Question 8
A glucose solution which boils at 101.04 ° C at 1 atm. What will be relative lowering of vapour pressure of an aqueous solution of urea which is equimolal to given glucose solution? (Given: $K_b$ for water is 0.52 K kg/mol)
Solution
$\Delta T_b= K_b.m$
$\Delta T_b = 101.04-100 = 1.04$ °C
Substututing the value of $\Delta T_b$ and $K_b$
or m= 1.04 /0.52 = 2 m
2 m solution means 2 moles of solute in 1 kg of solvent.
2 m aq solution of urea means 2 moles of urea in 1kg of water.
No. of moles of water = 1000/18 = 55.5
Now relative lowering of vapour pressure is given by the formula
Relative lowering of VP = x2 (where x2 is mole fraction of solute)
Relative lowering of VP = $n_2/n_1+n_2$ ($n_2$ is no. of moles of solute , $n_1$ is no. of moles of solvent)
= 2/ 2+55.5 = 2/57.5 = 0.034
Question 9
At 293 K vapour pressure of Ethyl acetate is 72.8 torr of Hg and that of ethyl propionate is 27.7 torr of Hg. assuming the mixture to be ideal find the V. P. off the mixture containing 25 g of Ethyl acetate and 50 g of Ethyl propionate.
Solution
for Ethyl acetate $CH_3COOC_2H_5$
$P_A^O=72.8$ torr
$W_A = 25$ gm , $M_A = 88$ gm/mol
For ethyl propionate $CH_3CH_2COO C_2 H_5$
$P_B^O = 27.7$ torr of Hg
$W_B = 50$ g,$M_B=102$ g/mol
$n_A = 25/88 = 0.284$
$n_B =50/102=0.490$
Now,
$\chi _A = n_A/(n_A+ n_B )=0.367$
$\chi _B = 1 – \chi _A = 1 – 0.367 = 0.633$
$P_S = .367 \times 72.8 + .633 \times 27.7=44.25$ torr of Hg
Question 10
Calculate the freezing point of a solution containing 0.5 g KCl (Molar mass = 74.5 g/mol) dissolved in 100 g water, assuming KCl to be 92% ionized.
$K_f$ of water = 1.86 K kg / mol.
Solution
KCL -> K
^{+} + Cl
^{-}
so, n=2
i=1-x + 2x = 1 +x
$\Delta T_f= iK_f.m$
$=.24$
$\Delta T_f=T_f^o- T_f$
$T_f=-.24 ^0$ C
Question 11
Benzene($C_6 H_6$) & Toulene($C_6 H_5 CH_3$) forms ideal solution. Vapour pressure of pure benzene & Toulene is 150 and 50 torr calc. the vapour pressure of the solution obtained when equal weights of the two substances are mixed.
Solution
$P_A^o = 150$ ,$P_B^o = 50$
$W_A$ = x,$W_B$ = x
$M_A = 78$,$M_B$ = 92
$n_A = x/78$,$n_B = x/92$
$x_A = n_A/(n_A+ n_B )=(x/78)/(x/78+x/92) = 46/85$
$x_B = 1- 46/85=39/85$
Total Vapour Pressure= = 104.1 torr.
Question 12
Calculate the freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be 90 % ionized.
[Given: Molar mass Br = 80 g/mol, $K_f$ of water = 1.86 K kg / mol]
Solution
HBr -> H
^{+} + Br
^{-}
so, n=2
i=1-x + nx = 1 +x
$\Delta T_f= iK_f.m$
$=3.53$
$\Delta T_f=T_f^o- T_f$
$T_f=-3.53 ^0$ C
Question 13
An aqueous solution comntaining 28% by weight of a liquid X (molecular mass = 140u), has a vapour pressure of 0.21 bar at 37° C. Find the vapour pressure of pure liquid (vapour pressure of water at 37° C is 0.198 bar)
Solution
$P_{water}^o = 0.198$ bar ,$P_x^o$= ?
$W_{water} = 72 $,$W_x = 28$
$M_{water} = 18 $,$M_x = 140$
$n_{water} = 72/18=4 $ ,$n_x = 28/140=0.2$
$x_{water} = 4/4.2$,$x_X = 0.2/4.2$
$P_S = P_x^o. x_X + P_{water}^o. x_{water}$
$0.210 = P_x^o (0.2/4.2)+ (0.198)(4/4.2)$
$P_x^o = 0.448$ bar
Question 14
What mass of a non – volatile solute urea should be dissolved in 100 gm of H2O to decrease the vapour pressure by 25%.
Answer 111 gm
Question 15The vapour pressure of Ethanol & meyhanol are 44.5 and 88.7 mm of Hg. An ideal solution formed by mixing 60 g of Ethanol & 40 g Methanol. Find the total vapour pressure the solution & the mole fraction of methanol vapour phase.
Question 16
Vapour pressure of pure benzene is 0.85 bar. A non volatile solute weighing 0.5 g is added to 39 g of benzene. The vapour pressure of the solution becomes 0.845 bar. Find the molar mass of solid substance.
Answer 170 g/mol
Question 17
Calculate the mass of a non – volatile solute (molar = 40 g/mol) which should be dissolved in 114 gm of octane to reduce its vapour pressure to 80 g
Answer 8 g
Question 18
Determine the Osmotic pressure of a solution which is prepared by dissolving 25 mg of $K_2SO_4$ in 2 L of water at 25° C.
Question 19
Calculate the vapour pressure of 1 M urea solution at 298 K the vapour pressure of pure water at 298 is 20 mm of Hg assume density of solution 1 gm per mm.
Answer 19.96 mm of Hg
Question 20
Find the value of Vant Haff Factor When the substance undergoes tetramerisation assuming
(a)complete association
(b)50% association
(c) 80% association
Solution
4x -> X4
(a) i = (no.of particles after association)/(no.of particles before association)
i = 1/4=0.25
(b) Incomplete association:
$\alpha = (i-1)/(n-1)$
$0.50 = (i-1)/(0.25 -1)$
So, $i= 0.725$
(c)
Incomplete association:
$\alpha = (i-1)/(n-1)$
$0.80 = (i-1)/(0.25 -1)$
So, $i= 0.4$
Question 21
Vant Haff facto for 0.1 M Barium Nitrate [$Ba(NO_3)_2$] is 2.74. Find the degree of dissociation.
Solution
n=3
$\alpha = (i-1)/(n-1)$
$\alpha =87$%
Question 22
The boiling point to benzene is 353.23 K. 1.8 gm of a non – volatile solute is dissolved in 19 gm of benzene. Boiling point is raised to 345.11 K. Find molal mass of solution. $K_b$ for $C_6H_6$ is 2.53 K kg.
Answer 58 g/mol
Question 23 In winter the normal temperature in a Himalayan valley is -10° C. Is 30% by mass of aqueous solution of ethylene glycol suitable for car radiators?
Solution
$W_A$ = 70 g,$W_B$ = 30 g
$M_B$ = 62 g
$K_f = 1.86$
$\Delta T_f= K_f \times (W_B \times 1000)/(M_B \times W_A )
$\Delta T_f= (1.86 \times 30 \times 1000)/(62 \times 70)
$=12$ ° C
$\Delta T_f= T_f^o- T_f $
$12=0- T_f $
$T_f = -12$ ° C.
As the freezing point is less than prevailing temperature.
Therefore, It is suitable to add.
Question 24
Find the Osmotic pressure of the solution which is obtained when 10 g of NaCl is present in 100 ml of solution. Assume that NaCl undergoes 80% dissociation at 27°C.
Question 25
What mass of NaCl should be dissolved in water to lower the freezing point by 7.5° C. $K_f$ for water is 1.86 K kg/mol. Assume that Vant Haff factor for NaCl is 1.87.
Answer is 8.2 g
Question 26
An aqueous solution of 1.248 g of $BaCl_2$ in 100 g of water boils at 100.0832 ° C. Kb for water is 0.52 K kg/mol. Find the degree of dissociation.Molar mass of $BaCl_2$ = 208 g/mol
Answer is 83.2%
Question 27
19.5g of Floroacetic acid ($CH_2FCOOH$) is dissolved in 500 g of water. The depression in freezing point is observed to be 1° C. Calculate Vant Haff factor & the dissociation constant of Floroacetic acid.
Question 28
The vapour pressure of pure water 25°C is 23.62 mm. What will be the vapour pressure of the solution containing 1.5 gm of urea in 50 gms of water?
Answer : 23.41 mm
Question 29
The osmotic pressure of a solutionb containing 85 g per litre of $AB_3$ (Molecular Mass 170) in water is 40 atm at 27° C. What is the degree of dissociation of $AB_3$
Answer is 75.1%
Question 30
Calculate the amount of NaCl whihc must be aded to 1Kg of water so that the freezing point by 3K. Given $k_f=1.86 K Kg/mol$
Answer is 47.2 g
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