Q: 3.42g of a substance of molecular weight 342, is present in 250g of water. Molality of this solution is:

Molality =\dfrac{\text{moles of solute}}{\text{mass of solvent}} \Rightarrow \dfrac{3.42}{342}\times \dfrac{1000}{250} \Rightarrow 0.04 m

Q: Molarity of 4\% (w/v) solution of NaOH is:

Molarity(M)=\cfrac {\text { number of moles of solute (n)}}{ \text {volume of solution (L)}} Step 1: Find moles of NaOH A per cent w/v solution is calculated with the following formula using the gram as the base measure of weight (w): % w/v = g of solute/ 100 mL of solutionWeight = 4 gm \text { Moles }=\dfrac{4\mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}} Step 2: Find MolarityAs we know, Molarity(M)=\cfrac {\text { number of moles of solute (n)}}{ \text {volume of solution (L)}} The molar mass of \mathrm{NaOH} is =40\ g/mol So, Molarity =\cfrac{4}{40} \times \cfrac{1000}{100}=1 \mathrm{M} So, The correct option is D .

Q: Calculate the volume in 10 millimoles of solute present in 0.08 M solution.

Molarity = \dfrac {Moles}{Volume (ml)}\times 1000 Volume = \dfrac {10\times 10^{-3}}{0\cdot 08}\times 1000 Volume = 125ml

Question 1

List - IList - II1) Solid in gasA) Brass2) Solid in solidB) Amalgam3) Liquid in solidC) Camphor in air4) Liquid in liquidD) Alcohol in waterWhich of the following is correct?

Accepted Answer

Camphor in the air is an example of a solution of solid in gas (camphor undergoes sublimation).Brass is an alloy of copper and tin(solid in solid).Amalgam is an example of a mixture of liquid in solid( Example  Zn  in mercury). Amalgam is a solid solution in which the solute is a liquid and the solvent is a solid.Whereas alcohol in water is an example of liquid in liquid( both are miscible).Therefore, option A is correct.

Question 2

Question 3

If air is taken as a binary solution, the solvent is:

Accepted Answer

(A) option is correct.Explanation:Air is mainly composed of oxygen and nitrogen gas. Around 78% is nitrogen gas and 21% is oxygen. Since the nitrogen content is more than oxygen so nitrogen will be the solvent and oxygen will be the solute if air is considered as binary solution.

Question 4

A sample of drinking water was found to be severely contaminated with chloroform,  {CHCl}_{3} , supposed to be a carcinogen. The level of contamination was  15  ppm (by mass). Express this in percent by mass.

Accepted Answer

15 ppm means  15  parts in million  (10^6)  parts \%\,by\,mass=\dfrac{15	imes 100}{10^{6}}=15	imes 10^{-4}                                              =1.5	imes 10^{-3} Hence, the correct option is  	ext{A}

Question 5

3.42g of a substance of molecular weight 342, is present in 250g of water. Molality of this solution is:

Accepted Answer

Molality =\dfrac{	ext{moles   of   solute}}{	ext{mass  of   solvent}} \Rightarrow \dfrac{3.42}{342}	imes \dfrac{1000}{250} \Rightarrow 0.04 m

Question 6

Molarity of  4\%  (w/v) solution of  NaOH  is:

Accepted Answer

Molarity(M)=\cfrac {	ext { number of moles of solute (n)}}{ 	ext {volume of solution (L)}} Step 1: Find moles  of  NaOH A per cent  w/v  solution is calculated with the following formula using the gram as the base measure of weight (w): % w/v  = g of solute/ 100  mL of solutionWeight =  4 gm 	ext { Moles }=\dfrac{4\mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}} Step 2: Find MolarityAs we know, Molarity(M)=\cfrac {	ext { number of moles of solute (n)}}{ 	ext {volume of solution (L)}} The molar mass of  \mathrm{NaOH}  is  =40\ g/mol So, Molarity  =\cfrac{4}{40} 	imes \cfrac{1000}{100}=1 \mathrm{M} So, The correct option is  D .

Question 7

Calculate the volume in 10 millimoles of solute present in 0.08 M solution.

Accepted Answer

Molarity = \dfrac {Moles}{Volume (ml)}	imes 1000 Volume = \dfrac {10	imes 10^{-3}}{0\cdot 08}	imes 1000 Volume = 125ml

Question 8

Question 9

H_{2}S , toxic gas with a rotten egg-like smell, is used for the qualitative analysis. If the solubility of  H_{2}S  in the water at STP is  0.195\ m . Calculate Henry's law constant.

Accepted Answer

Molality  =0.195m Solvent mass  =1kg Moles of solute  = 0.195 mol Number of moles of water  =\dfrac {1000g}{18}=55.56mol Molecular fraction of  H_2S=\dfrac {0.195}{0.195+55.5} \Rightarrow x=0.0035 Henry's constant  =K_H=\dfrac {P}{x}=\dfrac {0.987}{0.0035} [P=0.987\ atm,   at STP  ] K_H=282  atm

Question 10

Deep sea divers use  (O_2 + He)  mixture as preference to  (O_2 + N_2)  mixture. This is because:

Accepted Answer

Deep-sea divers prefer using the mixture of oxygen and helium to reduce the proportion of nitrogen and oxygen mixture. At high external pressure, the nitrogen mixed with blood, which then changes to toxic chemicals, which is fatal.However, in the case of helium and oxygen mixture, even if helium gets mixed with blood it will not cause any problem as helium is inert.

Question 11

Which characteristics the weak inter molecular forces of attraction in a liquid?

Accepted Answer

The stronger the intermolecular forces between the molecules, the harder it is for vaporization to occur since more energy is required to break the bonds holding the molecules together. So, weak intermolecular forces means high vapour pressure.

Question 12

If Raoult's law is obeyed, then vapour pressure of the solvent in a solution is directly proportional to:

Accepted Answer

According to Raoult's law; the partial vapour pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.For solvent: P=P_{A}^{o}X_A  ;  A\rightarrow  solvent P \ \propto \ X_A Hence, the correct option is  A .

Question 13

A mixture contains 1 mole of volatile liquid  A (P^{\circ}_{A}=100\ mm\ Hg)  and 3 moles of volatile liquid  B(P^{\circ}_{B}=80\ mm\ Hg) . If solution behaves ideally, the total vapour pressure of the distillate is :

Accepted Answer

Mole fraction of A is  X_A=\dfrac {1} {1+3}=0.25 The mole fraction of B is  X_B=1-0.25=0.75 The expression for the total pressure of the solution is  P=P^o_AX_A+P^o_BX_B Substitute values in the above expression P=100 	imes 0.25+80 	imes 0.75=85\  mm\ Hg Hence, the correct option is  A

Question 14

The vapour pressure of pure water is 23.5 mm Hg. Then, the vapour pressure of an aqueous solution which contains 5 mass percent of urea is:(Molar mass of urea is 60 )

Accepted Answer

Mass of urea  = 5\ g Mass of water  = 100 - 5 = 95\ g Number of moles of urea   = \dfrac {5}{60} = 0.083 Number of moles of water   = \dfrac {95}{18} = 5.278  	herefore  Total number of moles  = 5.278 + 0.083 = 5.361 Mole fraction of solvent  = \dfrac {5.278}{5.361} P_s =   Mole  fraction of solvent   	imes P^o = \dfrac {5.278}{5.361} 	imes 23.5         =  23.14 mm Hg   \approx   23 mm Hg

Question 15

Which of the following components form an ideal solution?

Accepted Answer

The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions.For ideal solution, the enthalpy of mixing of the pure components to form the solution is zero and the volume of  mixing is also zero, i.e., {	riangle }_{mix}H=0  and  {	riangle}_{mix} V=0 Thus type of solution is a solution of benzene and toluene.

Question 16

Which of the following condition is not followed for an ideal solution?

Accepted Answer

An ideal solution satisfies the following conditions.1)There will be no change in volume on mixing two components. 	riangle V_{mix}=0 2)There will be no charge in enthalpy on mixing two components. 	riangle H_{mix}=0 So, the  \Delta S_{mixing}=0  is not followed by the ideal solution.

Question 17

Match the column I with column II and mark the appropriate choiceColumn IColumn II(A)  { \Delta  }_{ mix }H=0,{ \Delta  }_{ mix }V=0 (i) Non-ideal solution(B)  { \Delta  }_{ mix }H
e 0,{ \Delta  }_{ mix }V
e 0 (ii) Positive deviation(C)  { \Delta  }_{ mix }H 0,{ \Delta  }_{ mix }V> 0 (iv) Negative deviation

Accepted Answer

(A)  { \Delta  }_{ min }H=0 ,   { \Delta  }_{ min }V=0    \longrightarrow   ideal solution(B)  { \Delta  }_{ mix }H
eq 0 ,   { \Delta  }_{ mix }V
eq 0    \longrightarrow   Non-ideal solution(C)  { \Delta  }_{ mix }H0 ,   { \Delta  }_{ mix }V>0    \longrightarrow   Positive deviation.

Question 18

A solution is obtained by dissolving 12g of urea (Mol.wt.=60) in a litre of solution. Another solution is prepared by dissolving 68.4g of cane sugar(Mol.wt.=342) in a litre of solution at the same temperature. The lowering of vapour pressure in the first solution is:

Accepted Answer

For urea  \Delta P_1=X_{urea}P^0 \Delta P_1=\dfrac {\dfrac {12}{60}}{total  \ mole}P^0 \Delta P_1=\dfrac {0\cdot 2}{total \ mole} P^0           ----(1)For sugar  \Delta P_2=X_{sugar}P^0 \Delta P_2=\dfrac {\dfrac {68\cdot 4}{342}}{total \ mole}P^0 \Delta P_2=\dfrac {0\cdot 2}{total \ mole} P^0           ----(2)Hence  (1)=(2) Hence, the correct option is  B

Question 19

200 mL  of an aqueous solution of a protein contains  1.26 g  of protein. The osmotic pressure of this solution at  300 K  is found to be  2.57 	imes {10}^{-3} bar  . The molar mass of protein will be: [R = 0.083  L bar mol ^{-1} K ^{-1} ]

Accepted Answer

The expression for the osmotic pressure of the solution is given below. \pi = CRT  Substitute values in the above expression. \pi= \displaystyle\frac { wt 	imes 1000 }{ GMM 	imes V } 	imes RT 2.57 	imes { 10 }^{ -3 } = \displaystyle\frac { 1.26 	imes 1000 }{ GMM 	imes 200 } 	imes 0.083 	imes 300 GMM = 61038  \ g Thus, the molar mass of protein is  61038\  g\  {mol}^{-1} Hence, the correct option is  	ext{D}

Question 20

Elevation in boiling point was  0.52^oC  when 6 g of compound X was dissolved in 100 g of water. Molecular mass of the compound M in  gmol^{-1}  is: ( K_b  for water is  0.52 Km^{-1} .)

Accepted Answer

We have \Delta T_b = K_b m 0.52 = 0.52 	imes \dfrac{mass\ of\ solute}{molar\ mass\ of\ solute\ 	imes mass\ of\ solvent\ (in\ g)} 	imes 1000 0.52 = 0.52 	imes \dfrac{6}{M	imes100}	imes1000 M =  60\ gmol^{-1}  Hence, option (B) is correct.

Question 21

A 0.5 M solution of weak electrolyte NaX at 285 K temperature is 10 % dissociated. What will be the osmotic pressure? ( \displaystyle R=0.082  L atm/mol /K)

Accepted Answer

\displaystyle NaX ightarrow Na^+ + X^-  Dissociation of 1 mole of NaX gives 2 moles of ions. So  \displaystyle n=2 The degree of dissociation  \displaystyle \alpha = 0.1   as solute is 10% dissociated. \displaystyle \alpha = \dfrac {i-1}{n-1} \displaystyle 0.1=\dfrac {i-1}{2-1} The Van't Hoff factor \displaystyle i=1.1 Now,Osmotic pressure  \displaystyle \Pi= iCRT = 1.1 	imes 0.5 	imes 0.082 	imes 285 = 12.9  atm.

Question 22

The molecular mass of acetic acid dissolved in water is  60  and when dissolved in benzene it is  120.  This difference in behaviour of  CH_3COOH  is because

Accepted Answer

Acetic acid (molar mass 60 g/mol) dimerizes in benzene due to the presence of hydrogen bonding between its molecule hence the molecular weight becomes double than that of normal value. while Water prevents the association of acetic acid.

Solutions