$$ = \dfrac {Normal \,moleular \,mass}{Observed \,molecular \,mass}$$

$$ = \dfrac {Observed \,molecular \,mass}{Normal \,molecular \,mass}$$

Less than one in case of dissociation

More than one in case of association

MCQ Questions for 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?

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

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?

0%
Relative lowering of vapour pressure
0%
Elevation of boiling point.
0%
Depression in freezing point
0%
Osmotic pressure

Mole fraction of A vapours above the solution in mixture of A and B (

$X_A=0.4$ )will be
[Given :

${ P }_{ A }^{ 0 }=100mmHg\quad and\quad { P }_{ B }^{ 0 }=200mmHg$ ]

0%
0.4
0%
0.8
0%
0.25
0%
none of these

Equimolar solutions of two non-electrolytes in the same solvent have:

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

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:

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?

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:

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:

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.

0%
True
0%
False

The van't Hoff factor (i) for an infinitely dilute solution of

$NaHSO_4$ is :

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 ?

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:

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:

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)

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.

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 :

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 :

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

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

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

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?

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?

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

$(Ag = 108)$

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

$litre^{-1}$ ?

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 H2O2sample, what mass of

$\mathrm{H}_{2} \mathrm{O}_{2}$ is there?

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

0%
Less than 1
0%
Greater than 1
0%
Equivalent to one
0%
Zero

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.

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

0%
78.2
0%
178.2
0%
206.2
0%
220

$0.5 g$ 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

$(K = 39, P = 31 )$

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

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?

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

0%
increases
0%
decreases
0%
remains the same
0%
depends upon climate

What per cent of the total aluminium in this ore is recoverable in the Bayer's process?

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?

$(F=19)$

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?

$(Cu=64)$

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 :

0%
$1 \%$
0%
$2 \%$
0%
$10 \%$
0%
$0.2 \%$

A normal solution is:

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

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:

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:

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

$=1.86^oC/m$ ]

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

0%
Unsaturated solution
0%
Saturated solution
0%
Both A and B
0%
Super saturated solution

Mass

$\%$ of carbon in ethanol is:

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 :

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

The vapour pressure of a liquid depends on

0%
Temperature but not on volume
0%
Volume but not on temperature
0%
Temperature and volume
0%
Neither on temperature nor 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

$\left(are\right)$

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 %

$H_2SO_4$ ?

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:

$R = 2 cal K^{-1} mol^{-1}$

0%
10 atm
0%
0.1 atm
0%
1 atm
0%
100 atm

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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