MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 8 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Solutions
Quiz 8
What will be the mass of glucose that would be dissolved in $$50$$ g of water in order to produce the same lowering of vapor pressure as is produced by dissolving $$1$$ g of urea in the same quantity of water?
Report Question
0%
$$1$$ g
0%
$$2$$ g
0%
$$3$$ g
0%
$$4$$ g
Which of the following sets of variables give a staright line with negative slope when plotted ?
( P = Vapour pressure ; T = Temperature in K)
y- axis x -axis y -axis x-axis
Report Question
0%
$$ P\quad \quad \quad T $$
0%
$$ log_{10}P\quad \quad \quad T $$
0%
$$ log_{10}P\quad \quad \quad \frac{1}{T} $$
0%
$$ log_{10}P \quad \quad \quad log_{10} \frac{1}{T} $$
Which of the following colligative properties can provide molar mass of proteins (or polymers orcolloids) with greater precision?
Report Question
0%
Relative lowering of vapour pressure
0%
Elevation of boiling point.
0%
Depression in freezing point
0%
Osmotic pressure
Explanation
Measurement of osmotic pressure provides a method to determine molar masses of solutes. This method is widely used to determine molar masses of proteins, polymers and other macromolecules.
Hence, option D is correct.
Mole fraction of A vapours above the solution in mixture of A and B ($$X_A=0.4$$)will be
[Given : $${ P }_{ A }^{ 0 }=100mmHg\quad and\quad { P }_{ B }^{ 0 }=200mmHg$$]
Report Question
0%
0.4
0%
0.8
0%
0.25
0%
none of these
Equimolar solutions of two non-electrolytes in the same solvent have:
Report Question
0%
same boiling point but different freezing point
0%
same freezing point but different boiling point
0%
same boiling and same freezing points
0%
Different boiling and differeent freezing points
Explanation
The elevation in the boiling point and the depression in the freezing point are $$colligative$$ properties and depend on the number of solute particles.
Equimolar solutions of non-electrolytes have the same number of solute particles. Hence, equimolar solutions of two non-electrolytes in the $$same$$ solvent have the same boiling point and same freezing point.
Hence, the correct option is C.
The relative lowering of vapour pressure of a solution containing $$6\ g$$ of urea dissloved in $$90\ g$$ of water is
Report Question
0%
$$0.0196$$
0%
$$0.05$$
0%
$$1.50$$
0%
$$0.01$$
6 g of urea is dissolved in 90 g of boiling water.The vapour preasure of the solution is:
Report Question
0%
$$744.8 mm$$
0%
$$758 mm$$
0%
$$761 mm$$
0%
$$760 mm$$
A solution of sodium sulphide, $$Na_2S$$ in water has a vapour pressure of $$17.5mm Hg$$ at a given temperature what is the mole fraction of $$Na_2S$$ in this solution if the vapour pressure of pure water is $$24.5mm Hg$$ at that temperature?
Report Question
0%
$$0.0476$$
0%
$$0.117$$
0%
$$0.143$$
0%
$$0.0953$$
The vapour pressure of two liquids are $$15000$$ and $$30000$$ in a unit. When equimolar solution of liquids is.. The the mole fraction of A and B in vapour phase will be:
Report Question
0%
$$\dfrac{2}{3}, \dfrac{1}{3}$$
0%
$$\dfrac{1}{3}, \dfrac{2}{3}$$
0%
$$\dfrac{1}{2}, \dfrac{1}{2}$$
0%
$$\dfrac{1}{4}, \dfrac{3}{4}$$
One mole of $${ N }_{ 2 }{ O }_{ 4 }(g)$$ at 100 K is kept in a closed container at 1.0 atm pressure. It is heated to 400 K , where 30% by mass of $${ N }_{ 2 }{ O }_{ 4 }(g)$$ decomposes of $${ N }O_{ 2 }(g)$$ .The resultant pressure will be:
Report Question
0%
4.2
0%
5.2
0%
3.2
0%
6.2
On diluting the solution, its relative lowering of vapour pressure changes but molality remains constant.
Report Question
0%
True
0%
False
Explanation
On diluting the solution, its relative lowering of vapour pressure changes
since it is a colligative property. But molality of the solution will also change
as
the mass of the solvent will change on dilution and molality depends on
the mass
of solvent too [molality = (moles of solute) / (mass of solvent in kg)]
The van't Hoff factor (i) for an infinitely dilute solution of $$NaHSO_4$$ is :
Report Question
0%
1/2
0%
1/3
0%
3
0%
2
The reflective lowering of sapour pressure of a solution containing $$6\ g$$ of urea dissolved in $$90\ g$$ of water is ?
Report Question
0%
$$0.0196$$
0%
$$0.05$$
0%
$$1.50$$
0%
$$0.01$$
The ratio by mass of the combining elements in the compound $$MgCl_2$$ is:
Report Question
0%
$$24:71$$
0%
$$14:3$$
0%
$$20:16$$
0%
$$12:3$$
Phenol associate in benzene as:
$${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons \cfrac{{\text{1}}}{{\text{2}}}{\left( {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}} \right)_{\text{2}}}$$
If degree of association of phenol is 40%. Van't Hoff factor is:
Report Question
0%
1
0%
0.8
0%
1.4
0%
0.6
The vapour pressure of a solution at $$100^{o}C$$ containing $$3\ g$$ of cane sugar in $$70\ g$$ of water? (vapour pressure of water at $$100^{o}C=760\ mm$$ of $$Hg$$) is)
Report Question
0%
$$758\ mm$$ of $$Hg$$
0%
$$730.36\ mm$$ of $$Hg$$
0%
$$400\ mm$$ of $$Hg$$
0%
$$700\ mm$$ of $$Hg$$
carbogen is...............
to a blood sample, ion present in hemoglobin is 3:72%. find no. of ion molecules in 2g of homoginity.
Report Question
0%
4.53$$\times { 10 }^{ 26 }$$
0%
4.53$$\times { 10 }^{ 23 }$$
0%
5.95$$\times { 10 }^{ 19 }$$
0%
8$$\times { 10 }^{ 20 }$$
$$P _ { A } = ( 235 y - 125 x y ) m m$$ of $$H g . P _ { A }$$ is partial pressure of A, $$x$$ is mole fraction of $$B$$ in liquld phase in the mixture of two liquids A and B and y is the mole fraction of $$A$$ in vapour phase, then
$${ P }^{ \circ }_{ B }$$
in mm of Hg is :
Report Question
0%
235
0%
0
0%
110
0%
125
A sample of air contains only $$\mathrm { N } _ { 2 } , \mathrm { O } _ { 2 }$$ and $$\mathrm { H } _ { 2 } \mathrm { O }$$ . It is saturated with water vapour and the total pressure is is $$640$$ $$torr$$. the vapour pressure of water is $$40$$ $$torr$$ and the molar ratio of $$\mathrm { N } _ { 2 } : \mathrm { O } _ { 2 } \mathrm { is } 3 : 1 .$$ The partial pressure of $$\mathrm { N } _ { 2 }$$ in the sample is :
Report Question
0%
$$480$$ $$torr$$
0%
$$600$$ $$torr$$
0%
$$525$$ $$torr$$
0%
$$450$$ $$torr$$
4 gm of a mixture of $${\text{CaC}}{{\text{O}}_3}$$ and sand is treated with an excess of HCl, and 0.88 gm of $${\text{C}}{{\text{O}}_2}$$ in $${\text{CaC}}{{\text{O}}_3}$$ the original mixture will be
Report Question
0%
20%
0%
50%
0%
25%
0%
38%
$$ O .8 M F \mathrm{eSO}_{4} $$ solution requires 160 $$ \mathrm{ml}$$ , 0.2 $$\mathrm{M}$$ $$A I_{2}\left(C r_{2} O_{7}\right)_{3}$$ in acidic medium, Calculate volume
Report Question
0%
480 ml
0%
240 ml
0%
720 ml
0%
40 ml
In a biogas plant $${ CH }_{ y }$$ is produced as fuel find the mass ratio of carbon to hydrogen
Report Question
0%
3 : 1
0%
2 : 1
0%
1 : 3
0%
None
One litre of a mixture of $$ O_{2}$$ and $$ O_{3}$$ at $$ 0^{\circ}C $$ and $$1 atm$$ was allowed to react with an excess of an acidified solution of $$KI$$. The iodine liberated requires $$40 ml$$ of $$M/10$$ sodium thiosulphate solution for titration. What is the mass per cent of ozone in the mixture?
Report Question
0%
$$6.575$$%
0%
$$9.6$$%
0%
$$93.425$$%
0%
$$90.4$$%
Explanation
Let the total moles of $$O_2$$ and $$O_3$$ in the mixture be n.
Applying $$PV=nRT$$
$$1×1=n×0.0821×273$$
$$n=0.044\ mole$$
Moles of $$O_3$$= Moles of $$I_2$$=$$\dfrac{1}{2}$$ moles of $$Na_2S_2O_3$$
$$=\dfrac{1}{2}×\dfrac{1}{10}×\dfrac{40}{1000}=0.002\ moles$$
Moles of $$O_2$$ in the mixture = $$0.044−0.002=0.042\ moles$$
Mass of $$O_2=0.042×32=1.344g$$
Mass of $$O_3=0.002×48=0.096g$$
Total Mass of mixture $$=1.344+0.096=1.440$$
Mass% $$O_3=\dfrac{0.096}{1.440}×100=6.575$$%
Hence, option $$(A)$$ is a correct answer.
Which of the following sets of components form homogenous mixture?
Report Question
0%
Phenol$$+$$Water
0%
Sugar$$+$$Benzene
0%
Silver chloride$$+$$Water
0%
Ethyl alcohol$$+$$Water
Explanation
The homogenous mixture has all the components in the same phase.
Solubility of alcohols and phenols in
water is due to their ability to form
hydrogen bonds with water molecules
. The solubility decreases with
increase in size of alkyl/aryl (hydrophobic) groups. Several of the lower
molecular mass alcohols (ethyl alcohol) are miscible
with water in all proportions.
Phenol+ water is less homogeneous than ethyl alcohol+water beacause of bigger size of aryl group in phenol.
$$\therefore$$ option D is correct.
The normality of a solution of a mixture containing $$HCl$$ and $$ H_{2}SO_{4}$$ is $$N/5$$. Twenty millilitres of this solution reacts with an excess of $$ AgNO_{3}$$ solution to give $$0.287 g$$ of silver chloride. The percentage of $$HCl$$ in the mixture by mass, is $$(Ag = 108)$$
Report Question
0%
$$42.69$$%
0%
$$57.31$$%
0%
$$50$$%
0%
$$25$$%
Explanation
no. of equivalent in the mixture $$ = 0.02\times (\dfrac{1}{5}) $$
$$ = 0.004\,equt $$
no. of equt of $$HCl$$ = no. of equt of $$AgCl$$
no. of equt $$HCl$$ $$ = \dfrac{0.287}{143.5} = 0.002 $$
no. of. equt of $$ H_{2}SO_{4} $$ in mixture $$ = 0.004-0.002 $$
$$ = 0.002 $$
(eqnt at of $$ H_{2}SO_{4} = 49 $$)
% by mess of $$HCl$$ in mixture
$$ = \dfrac{0.002\times 36.5}{0.002\times 36.5+0.002\times 49}\times 100 $$
$$ = 42.69\% $$
ans Option (a)
Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of $$1$$ atm, also a concentration of $$1$$ mole $$litre^{-1}$$?
Report Question
0%
At $$STP$$
0%
When $$V=22.4$$
0%
When $$T=12K$$
0%
Impossible under any condition
Explanation
$$PV=nRT$$ Ideal gas equation
$$P=\dfrac{n}{V}RT$$
Concentration$$=\dfrac{n}{V}=1$$
$$P=1$$ atm
$$R=0.0821$$
$$1=1(0.0821)(T)$$
$$T=\dfrac{1}{0.0821}=12K$$
$$\therefore T=12K$$
Option C.
In 300 mL of a 5 volume H2O2
sample, what mass of $ \mathrm{H}_{2} \mathrm{O}_{2} $ is there?
Report Question
0%
18.2 g
0%
9.1 g
0%
4.55 g
0%
none of these
The experimental molecular weight of an electrolyte will always be less than its calculated value bacause the value of Van't Hoff factor "i " is
Report Question
0%
Less than 1
0%
Greater than 1
0%
Equivalent to one
0%
Zero
Explanation
The experimental molecular weight of an electrolyte will always be less than its calculated value bacause the value of i is greater than unity (i.e. i > 1), for dissociation.
The vapour pressure of pure $$M$$ and $$N$$ are $$700$$mm of $$Hg$$ and $$450 mm$$ of $$Hg$$ respectively.which of the following option is correct?
Given :$$X_N,X_M$$- mole fraction of $$N$$ and $$M$$ in liquid phase
$$Y_N,Y_M$$= mole fraction of $$N$$ and $$M$$ in vapour phase.
Report Question
0%
$$X_M-X_N>Y_M-Y_N$$
0%
$$\dfrac{X_M}{X_N}>\dfrac{Y_M}{Y_N}$$
0%
$$\dfrac{X_M}{X_N}<\dfrac{Y_M}{Y_N}$$
0%
$$\dfrac{X_M}{X_N}=\dfrac{Y_M}{Y_N}$$
Explanation
Solution:- (C) $$\dfrac{X_M}{X_N}<\dfrac{Y_M}{Y_N}$$
Mole fraction of more voltatile component increase in vapour phase, i.e.,
$$P_N^o<P_M^o$$
The vapours pressure of benzene at $$90^oC$$ is 1020 torr A solution of 5 g of a solute in 58.5 g benzene has vapour pressure 990 torr the molecular weight of the solute is
Report Question
0%
78.2
0%
178.2
0%
206.2
0%
220
A $$0.5 g$$ sample of $$ KH_{2}PO_{4} $$ is titrated with $$0.1 M$$ $$NaOH$$. The volume of base required to do this is $$25.0 ml$$ The reaction is represented as :
$$ H_{2}PO_{4} ^{-}+OH ^{-}\rightarrow HPO_{4}^{2-}+H_{2}O $$.
The percentage purity of $$ KH_{2}PO_{4} $$ is
$$(K = 39, P = 31 )$$
Report Question
0%
$$68$$%
0%
$$34$$%
0%
$$85$$%
0%
$$51$$%
Explanation
moles of Base $$= 0.1 \times \dfrac{25}{1000}$$
$$ = 2.5 \times 10^{-3}$$
$$ \therefore $$ moles of $$ KH_{2}PO_{4} \equiv 2.5 \times 10^{-3}$$
$$ \therefore $$ moles of $$ KH_{2} PO_{4} = 2.5 \times 10^{-3} \times 136$$
$$ = 0.34 \, gram $$
% purity $$ = \dfrac{0.34}{0.50} \times 100 $$
= 68%
option "a" correct.
What is the highest percentage yield of $$N_{2}$$ that can be expected? The theortical yield the quantity of $$N_{2}$$ formed in the absence of parallel reaction
Report Question
0%
$$96.67\%$$
0%
$$90\%$$
0%
$$85.7\%$$
0%
$$100\%$$
Explanation
$$2N_{2}H_{4}+N_{2}O_{4}\rightarrow 3N_{2}+4H_{2}$$
Mole of $$N_{2}H_{4}=\dfrac{95}{52}=3$$ mole
Mole of $$N_{2}O_{4}=\dfrac{184}{92}=2\ mole$$
$$2N_{2}H_{4}=1 N_{2}O_{4}$$
$$3-1=2$$
$$x=\dfrac{3}{2}=1.5\ mole$$
limiting reactant $$N_{2}H_{4}$$
Mole of $$NO=\dfrac{18}{30}=6\ mole$$
Paralle ray $$N_{2}H_{4}+N_{2}O_{4}\rightarrow 2NO+N_{2}+2H_{2}O --- (2)$$
$$1N_{2}H_{4}=2NO$$
$$x-1=0.6\ NO$$
$$x=0.3\ mole$$
actual mole of $$N_{2}H_{4}$$ available in $$Pxn 1=3-0.3=2.7\ mole$$
$$N_{2}$$ produced in $$Rxn 1=\dfrac{2.7\times 3}{2}=4.05\ mole$$
$$N_{2}$$ Produced in $$Rxn =\dfrac{0.3\times 1}{7}=0.3\ mole$$
$$(N_{2})$$ Total $$=4.05+0.3=4.35\ mole$$
But actual tield of $$N_{2}=\dfrac{3\times 3}{2}=4.5\ mole$$
$$\% N_{2}=\dfrac{4.35}{4.5}\times 100=96.67\%$$ Option $$(a)$$
What is the percentage of copper in brass?
Report Question
0%
$$80\%$$
0%
$$90\%$$
0%
$$85\%$$
0%
$$10\%$$
Explanation
Balanced eqn is
$$Cu+4HNO_{3}\rightarrow Cu(NO_{3})_{2}+2NO_{2}+H_{2}O$$
$$4Zn+10HNO_{3}\rightarrow NH_{4}HO_{3}+4Zn (NO_{3})_{2}+3H_{2}O$$
Mole of $$HO_{2}=\dfrac{PV}{RT}=\dfrac{1\times 1.04}{0.0821\times 298}=0.0425\ mole$$
Mole of $$Cu=0.0212\times 63.5=1.346\ g$$
$$\% Cu=\dfrac{1.346}{1.5}\times 100=89.74\% \approx 90\%$$
Option $$(b)$$
On increasing the altitude at constant temperature, vapour pressure of liquid
Report Question
0%
increases
0%
decreases
0%
remains the same
0%
depends upon climate
Explanation
Vapour pressure
is a
temperature
dependent phenomena and it does not change unless
temperature
is changed.
On increasing altitude
, the outside
pressure
decreases and to maintain that the rate of evaporation and condensation also changes which keeps
vapour pressure constant
. So answer is option C.
What per cent of the total aluminium in this ore is recoverable in the Bayer's process?
Report Question
0%
$$80$$
0%
$$90$$
0%
$$85$$
0%
$$75$$
Explanation
Mole wt of kaolin $$=2589/mole$$
Mole of $$Al$$ is kaolin $$=\dfrac{13}{258}=0=5$$
$$(Al_{2}O_{3})$$
Mole wt of Gibbrite $$=156 9/mole$$
Mole of gibbrite $$=\dfrac{87}{156}=0.557\ mole$$
Total mole of $$Al_{2}O_{3}=0.557+0.05=0.607\ mole$$
Mole of $$HO_{2}$$ from kaolin $$=2\times 0.05=0.01\ mole$$
Max mole of Muld forms $$=\dfrac{0.1}{5}=0.02\ mole$$
$$1$$ mole Mud $$=3$$ mole $$N_{2}O_{3}$$
$$0.02$$ mole Mud $$=x-1$$
$$x=3\times 0.02=0.06$$
Max $$Al_{2}$$ can be expected $$=0.607-0.06=0.547$$
$$\% Al$$ can be expected $$=\dfrac{0.547}{0.607}\times 100=90.11\%$$
Option $$(b)$$
When a sample of hydrogen fluoride is cooled to $$303\ K$$, most of the molecules undergo dimerization. If the vapour density of such a sample is $$18$$, what percent of total molecules in the sample are in dimer from? $$(F=19)$$
Report Question
0%
$$88.89$$
0%
$$80.0$$
0%
$$20.0$$
0%
$$11.11$$
Explanation
$$mw=$$ vapour density $$\times 2=16$$
$$HO\to (HP)_2$$
Suppose $$1$$ mole of $$HF$$
$$20X+40(-X)=36$$
$$20X+40-40X=36$$
$$-20X=-4$$
$$X=\dfrac {1}{3}r=2$$
$$\dfrac {4}{5}\times 100=80\%$$
A sample of pure $$Cu (4.00\ g)$$ heated in a stream of oxygen for some time, gains in weight with the formation of black oxide of copper $$(CuO)$$. The final mass is $$4.90\ g$$. What percent of copper remains unoxidzed? $$(Cu=64)$$
Report Question
0%
$$90\%$$
0%
$$10\%$$
0%
$$20\%$$
0%
$$80\%$$
Explanation
$$2Cu+O_2\to 2CuO$$
$$2\times 64\ gm\ Cu$$ reacts with $$32\ gm\ O_2$$ to form $$160\ gm\ CuO$$
$$=128\ gm\ Cu$$
Initially $$4\ gm\ Cu\quad O\ gm$$
Final $$4-n\ gm$$
$$\therefore \ n\ gm\ Cu$$ form $$=\dfrac {160}{128}\times x\ gm\ CuO$$
$$4-x+=\dfrac {160x}{128}=4.90$$
$$\therefore \ x=3.6\ gm$$
$$\therefore \ Cu$$ unrected $$=4-x=4-3.6=0.4\ gm$$
$$\%$$ of $$Cu$$ unoxidised $$=\dfrac {0.4}{4}\times 100=10\%$$
The percentage (by volume) of argon in the atmosphere is about :
Report Question
0%
$$1 \%$$
0%
$$2 \%$$
0%
$$10 \%$$
0%
$$0.2 \%$$
A normal solution is:
Report Question
0%
one gram equivalent mass of the substance in one litre solution
0%
one gram molecular mass of the substance in one litre solution
0%
one gram equivalent mass of the substance in mL of the solution
0%
is that whose concentration is known
Explanation
A normal solution contains one equivalent of solute per litre of solution.
For example, a mole of $$HCl$$ or $$NaOH$$ is one equivalent, but a mole of $$H_2SO_4$$ or $$Ca(OH)_2$$ is two equivalents.
Thus, a six-molar (6 M) sulfuric acid solution is twelve-normal (12 N).
Van't Hoff factor i
Report Question
0%
$$ = \dfrac {Normal \,moleular \,mass}{Observed \,molecular \,mass}$$
0%
$$ = \dfrac {Observed \,molecular \,mass}{Normal \,molecular \,mass}$$
0%
Less than one in case of dissociation
0%
More than one in case of association
Explanation
The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.
$$=\dfrac{\text{normal molecular mass}}{\text{observed molecular mass}}$$
Hence, the correct option is A.
When 20 g of naptholic acid $$ ( C_{11}H_8O_2) $$ is dissolved in 50 g of benzene $$ (K_f = 1.72 K kg mol^{-1} ) $$. A freezing point depression of 2 K is observed. The van't hoff factor (i) is:
Report Question
0%
0.5
0%
1
0%
2
0%
3
Explanation
$$\mathbf{Van'tHoff\ factor(i)=\cfrac{actual\ mol.\ wt.}{calculate\ mol.\ wt.}}\rightarrow (1)$$
The actual molecular weight of naphthoic acid $$(C_{11}H_{8}O_{2})=172\rightarrow (2)$$
Calculated molar mass$$=\cfrac{1000.w.K_{f}}{W. \Delta T_{f} }$$
where,
$$\Delta T_{f}=$$Change in freezing point$$=2K$$
$$i=$$Van't Hoff factor$$=?$$
$$K_{f}=$$Freezing point constant$$=1.72Kgmol^{-1}$$
$$w(weight\ of\ naphthoic\ acid)=20g$$
$$W(weight\ of\ benzene)=50g$$
$$=\cfrac{1000\times 1.72\times 20}{50\times 2}$$
$$=344$$
calculated molecular weight of Naphthoic acid$$ = 344\rightarrow (3)$$
From equation $$(1),(2),(3)$$
$$Van't Hoff\ factor(i)=\cfrac{actual\ mol.\ wt.}{calculate\ mol.\ wt.}=\cfrac{172}{344}=0.5$$
$$\mathbf{Hence\ the\ correct\ answer\ is\ option\ (A)}$$
Mass percentage of deuterium in heavy water is:
Report Question
0%
same as that of protium in water
0%
$$11.1$$
0%
$$20.0$$
0%
cannot be predicted
Explanation
Heavy water$$(D_2O)$$
is also called deuterium oxide.
Deuterium, the hydrogen isotope has a mass double of that of ordinary hydrogen.
The molecular weight of
$$(D_2O)$$
=2(2)+16=20 g/mol
The percentage weight of deuterium in heavy water is $$=\dfrac{4}{20}×100=20$$%
Hence, option $$C$$ is correct.
A complex is prepared by mixing $$CoCl_3$$ and $$NH_0.1\ M$$ solution of the complex was found to freeze at $$0.372 ^oC$$. The formula of the complex is :
[ Molal depression constant of water $$=1.86^oC/m$$]
Report Question
0%
$$[Co(NH_3)_6]Cl_3$$
0%
$$[Co(NH_3)_5Cl]Cl_2$$
0%
$$[Co(NH_3)_4Cl_2]Cl$$
0%
$$[Co(NH_3)_3Cl_3]$$
Explanation
Theoretical value of
$$ΔT_f=K_f × m =1.86^oC/m × 0.1 m = 0.186$$
Observed value of $$ΔT_f= 0.372$$
As observed $$ΔT_f$$ is twice the theoretical value, this shows that each molecule of the complex dissociates to form two ions. This can be possible only if the formula of the complex is $$[Co(NH _3)_ 4Cl_ 2]Cl$$
In a glass we are having half filled water. Then we are continuously adding salt to the water and salt is getting dissolved into water. But a state occur when water is not able to dissolve salt. Then the solution is called as
Report Question
0%
Unsaturated solution
0%
Saturated solution
0%
Both A and B
0%
Super saturated solution
Explanation
A saturated solution is formed due to maximum concentration of a solute dissolved into solvent. The additional solute will not dissolve in a saturated solution.
Mass $$\%$$ of carbon in ethanol is:
Report Question
0%
$$52$$
0%
$$13$$
0%
$$34$$
0%
$$90$$
0%
$$80$$
Explanation
Molecular mass of $$C_2H_5OH=46.00$$
$$\because\begin{bmatrix} Atomic & mass & of & C=12.00\\ Atomic & mass & of & O=16.00\\ Atomic & mass & of & H=1.00\end{bmatrix}$$
Also,
$$\because 46.00$$ of $$C_2H_5OH$$ contain, $$C=24$$g
$$\because 100$$g of $$C_2H_5OH$$ contain, $$C=\dfrac{24\times 100}{46}= 52.17\%\equiv52\%$$.
A molal situation is one that contains one mole of the solute in :
Report Question
0%
1000 g of the solvent
0%
one litre of the solvent
0%
one litre of the solution
0%
22.4 litre of the solvent
Explanation
Molality is the number of moles of solute dissolved in 1000 grams of solvent.
So 1 molal solution contains one mole of solute in 1000 grams of solvent.
Hence, option A is correct.
The vapour pressure of a liquid depends on
Report Question
0%
Temperature but not on volume
0%
Volume but not on temperature
0%
Temperature and volume
0%
Neither on temperature nor on volume
Explanation
When the vapour pressure of solvent decreases, then the boiling point of solvent increases. The vapour pressure of a liquid depends on temperature but not on volume.
For a solution formed by mixing liquids $$L$$ and $$M$$, the vapour pressure of $$L$$ plotted against the mole fraction of $$M$$ in solution is shown in the following figure. Here $$x_L$$ and $$x_M$$ represent mole fractions of $$L$$ and $$M$$ respectively in the solution. The correct statement $$\left(s\right)$$ applicable to this system is $$\left(are\right)$$
Report Question
0%
The point $$Z$$ represents vapour pressure of pure liquid $$M$$ and Raoults law is obeyed from $$X_{L}=0$$ to $$X_{L}=1$$.
0%
Attractive intermolecular interactions between $$L-L$$ in pure liquid $$L$$ and $$M - M$$ in pure liquid $$M$$ are stronger than those between $$L - M$$ when mixed in solution.
0%
The point $$Z$$ represents vapour pressure of pure liquid $$M$$ and Raoults law is obeyed when $$X_L \to 0$$.
0%
The point $$Z$$ represents vapour pressure of pure liquid L and Raoults law is obeyed when $$X_L\to 1$$
Explanation
We know from Raoult's law,
$$P=P_L^ox_L+{P_M}^ox_M$$
Vapour Pressure of liquid,
$$P_L={P_L}^ox_L=P_L^o(1-x_M)$$ at $$x_M\rightarrow 0$$ or $$x_L \rightarrow 1$$
$$\therefore P_L={P_L}^o$$ This follows the graph at point Z
Option D is correct.
The statement B is also correct because the attraction in like molecules are more than that in unlike molecules.
Option B is also correct.
Ans- B, D.
Oleum is considered as a solution of $$SO_3$$ in $$H_2SO_4$$. Which is obtained by passing $$SO_3$$ in solution of $$H_2SO_4$$.When 100 g sample of oleum is diluted with desired weight of $$H_2O$$ then the total mass of $$H_2SO_4$$.obtained after dilution is known as labelling of oleum.
For example , a oleum is diluted by 9 g of $$H_2O$$ which combined with all the free $$SO_3$$ to form $$H_2SO_4$$ as:
$$SO_3 +H_2O \rightarrow H_2SO_4$$.
What is the % of free $$SO_3$$ in an oleum that is labelled as 104.5 % $$H_2SO_4$$?
Report Question
0%
10
0%
20
0%
40
0%
None of these
Explanation
$$SO_3 + H_2O \rightarrow H_2SO_4$$
18 g water combines with 80 g $$SO_3$$
$$\therefore$$ 4.5 g of $$H_2O$$ combines with 20 g of $$SO_3$$
$$\therefore$$ 100 g of oleum contains 20 g of $$SO_3$$
or, 20% free $$SO_3$$.
$$\Delta_f G^{\circ}$$ at 500 K for substance 'S' in liquid state and gaseous state are +100.7 kcal $$mol^{-1}$$ respectively. Vapour pressure of liquid 'S'' at 500 K is approximately equal to:
$$R = 2 cal K^{-1} mol^{-1}$$
Report Question
0%
10 atm
0%
0.1 atm
0%
1 atm
0%
100 atm
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 12 Engineering Chemistry Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
