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451.

Explain the elevation in the boiling point of a solution.

Answer»

The elevation in the boiling point of a solution is defined as the difference between the boiling points of the solution and the pure solvents at a given pressure, e.g. If T0 and T are the boiling points of a pure solvent and a solution, then the elevation in boiling point, ΔTb = T – T0. It is a colligative property.

452.

For an ideal binary liquid solutions with `P_(A)^(@)gtP_(B)^(@)`, which relation between `X_(A)` (mole fraction of `A` in liquid phase) and `Y_(A)` (mole fraction of `A` in vapour phase) is correct:A. `"Y"_("A")lt"Y"_("B")`B. `"X"_("A")gt"X"_("B")`C. `("Y"_("A"))/("Y"_("B"))gt("X"_("A"))/("X"_("B"))`D. `("Y"_("A"))/("Y"_("B"))lt("X"_("A"))/("X"_("B"))`

Answer» Correct Answer - C
453.

For an ideal binary liquid solutions with `P_(A)^(@)gtP_(B)^(@)`, which relation between `X_(A)` (mole fraction of `A` in liquid phase) and `Y_(A)` (mole fraction of `A` in vapour phase) is correct:A. `X_(A)=Y_(A)`B. `X_(A)gtY_(A)`C. `(X_(A))/(X_(B))lt(Y_(A))/(Y_(B))`D. `X_(A),Y_(A),X_(B)` and `Y_(B)` cannot be correlated

Answer» Correct Answer - C
Mole fraction more volatile substance is greater in vapour phase.
454.

For an ideal binary liquid solution with `p_(A)^(@) gt p_(B)^(@)` which of the following relations between `x_(A)` (mole fraction of A in liquid phase) and `y_(A)` (mole fraction of A in vapour phase) is correctly represented?A. `x_(A) = y_(A)`B. `x_(A) gt y_(A)`C. `x_(A) lt y_(A)`D. `x_(A)` and `y_(A)` cannot be correlated with each other

Answer» Correct Answer - C
`y_(A) = (p_(A))/(P_("Total")) = (x_(A)p_(A)^(@))/(P_("Total"))` or `(y_(A))/(x_(A)) = (p_(A)^(@))/(P_("Total"))`
As B is less volatile than `A, P_("Total") lt p_(A)^(@)`
`:. (y_(A))/(x_(A)) gt 1` or `y_(A) lt x_(A)`
455.

Solve the following :Henry’s law constant for the solubility of methane in benzene is 4.27 × 10-5 mm Hg mol dm-3 at constant temperature. Calculate the solubility of methane at 760 mm Hg pressure at same temperature.

Answer»

Given :

Henry’s law constant = K 

= 4.27 × 10-5 mm Hg mol dm-3 

Pressure of the gas = P = 760 mm Hg

KH = 4.27 × 10-5 mm-1 mol dm-3

= 4.27 × 10-5 × 760 atm-1 mol dm-3

= 3245 × 10-5 atm-1 mol dm-3

P = 760 mm = \(\frac{760}{760}\) = 1 atm

By Henry’s law,

S = KH x P = 3.245 x 10-2atm-1 mol dm-3 x 1 atm

= 3.245 × 10-2 mol dm-3

∴ Solubility of methane = 3.245 × 10-2 mol dm-3

456.

STATEMENT-1 : Molality and mole fraction are temperature independent quantity. and STATEMENT-2 : Molality and mole fraction are unit less quantity.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-5B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True

Answer» Correct Answer - 3
457.

Out of moalrity (M), molality (m), formality (F) and mole fraction (x) those independent of temperature are:A. M,mB. F,xC. m,xD. M,x

Answer» Correct Answer - C
Molality, m and mole fraction x are based on mass-mass relationship. Volume is nto involved in them.
458.

(a) Explain the following phenomena with the help of Henry’s law.(i) Painful condition known as bends.(ii) Feeling of weakness and discomfort in breathing at high altitude.(b) Why soda water bottle kept at room temperature fizzes on opening?

Answer»

(a)
(i) Deep sea divers depend upon compressed air for breathing at high pressure under water. The compressed air contains  N2 in addition to O2, which are not very soluble in blood at normal pressure. However, at great depths when the diver breathes in compressed air from the supply tank, more N2 dissolves in the blood and other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. When the diver comes towards the surface, the pressure decreases, N2 comes out of the body quickly forming bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of nerve impulses. The bubbles can even burst the capillaries or block them and starve the tissues of O2. This condition is called the bends, which are painful and life-threatening.
(ii) At high
altitudes the partial pressure of O2 is less than that at the ground level. This results in low concentration of oxygen in the blood and tissues of the people living at high altitudes or climbers. The low blood oxygen causes climbers to become weak and unable to think clearly known as anoxia.
(b) To increase the solubility of CO2 in soft drinks, the soda water bottles are sealed under high pressure. When the bottle is opened at room temperature under normal atmosphere conditions, the pressure inside the bottle decreases to atmospheric pressure and excess CO2 fizzes out.

Refer to the NCERT textbook for Class XII.
459.

Explain the solubility rule “like dissolves like” in terms of intermolecular forces that exist in solutions.

Answer»

A substance (solute) dissolves in a solvent if the intermolecular interactions are similar in both the components; for example, polar solutes dissolve in polar solvents and non polar solutes in non polar solvents thus we can say “like dissolves like”.

460.

Consider the Fig. 2.1 and mark the correct option. (i) water will move from side (A) to side (B) if a pressure lower than osmotic pressure is applied on piston (B). (ii) water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B). (iii) water will move from side (B) to side (A) if a pressure equal to osmotic pressure is applied on piston (B). (iv) water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A).

Answer»

(ii) water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B).

461.

Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.

Answer»

Molarity of a solution is defined as the number of moles of solute dissolved in one litre of solution. Since volume depends on temperature and undergoes a change with change in temperature, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, as a result other concentration terms given in the question remain unchanged by changing temperature. According to the definition of all these terms, mass of the solvent used for making the solution is related to the mass of solute.

462.

Why are aquatic species more comfortable in cold water in comparison to warm water?

Answer»

At a given pressure the solubility of oxygen in water increases with decrease in temperature. Presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

463.

What is the significance of Henry’s Law constant KH?

Answer»

Higher the value of Henry’s law constant KH, the lower is the solubility of the gas in the liquid.

464.

The value of Henry’s constant KH is _____________.(i) greater for gases with higher solubility.(ii) greater for gases with lower solubility.(iii) constant for all gases.(iv) not related to the solubility of gases.

Answer» (ii) greater for gases with lower solubility.
465.

Value of Henry’s constant KH ____________.(i) increases with increase in temperature.(ii) decreases with increase in temperature.(iii) remains constant.(iv) first increases then decreases.

Answer» (i) increases with increase in temperature.
466.

3.4 g H2O2 is dissolved in 448 mL solution. Thesolution will be labelled as (V = Vol. strength):-A) 0.224VB) 5VC) 2.5VD) 20V

Answer»

Strength of H2O2 = \(\frac{Weight\,of\,H_2O_2}{Volume\,of\,solution(L)}\)

\(\frac{3.4g\times1000}{448L}\)

= 7.6 g/L
\(\underset{(2\times34g)}{2H_2O_2}\longrightarrow2H_2O+\underset{(22.4L)\,at\,NTP}{O_2}\) 

68 g of H2O2 produces 22.4 litres at oxygen gas at NTP.

Therefore, 

7.6g of H2O2 will produce

= 22.4/68 x 7.6

= 2.5 litre at oxygen.

Thus, volume strength of given H2O2 solution is 2.5

Hence, option(c) is Right answer.

467.

Will osmosis take place when 0.1 M aqueous urea and glucose solutions are separated by semipermeable membrane ?

Answer» No, it will not take place as they are isotinic as well as iso-osmotic in nature.
468.

If `0.1 M` solution of glucose and `0.1 M` solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be be correct to say thatA. water will flow from urea solution to glucoseB. urea will flow toward glucose solutionC. glucose will flow toward urea solutionD. there will be no net movement

Answer» Correct Answer - 4
Since 0.1 M solution of glucose and 0.1 M solution of urea an isotonic to each other, they will have the same osmotic pressure at the same temperature. Hence, there will be no net movement across the semipermeable membrane.
469.

Aqeous solution of which acid is called "stomach acid"?A. Acetic acidB. Oxalic acidC. Hydrochloric acidD. Sulphuric acid

Answer» Correct Answer - 3
Hydrocholoric acid, `HCl(aq.)`, is called "stomach acid" because it is the main acid `(~ 0.10 M)` in our digestive juices. Note that digestive juice is the acidic fluid secreted by glands in the lining of the stomach. When the concentration of `HCl` is too high in humans, problems result. These problem may range from "heartburn" to ulcers that can eat through the lining of the stomach wall.
Snakes have very high concentrations of HCl in their digestive juices so that they can digest whole small animals and bbirds
Automobile batteries contain `40% H_(2)SO_(4)` by mass. When the battery has "ran down" the concentration of `H_(2)SO_(4)` is significantly lower than `40%`. A technician checks an automobile battery by drawing some acid into a hydrometer, which indicates the density of the solution. This density is related to the concentration of `H_(2)SO_(4)`.
Oxalic acid `(COOH)_(2)` is used to remove rust and some ink stains from fabrics.
470.

What is the fraction of the solute in a `1.00` m aqueous solution ?A. 1.77B. 0.177C. 0.0177D. 0.0344

Answer» Correct Answer - 3
`1.00` molal aqueous solution means `1 mol` of the solute is present in enery `1 kg (1000 g)` of solvent Considering `1 kg` of solvent, we have
`n_(solute)=1`
`n_(solvent)=(m_(H_(2)O))/(mm_(H_(2)O))=(1000g)/(18 g//mol)`
`=55.6 mol`
Thus, `chi_(solute)=n_(solute)/(n_(solute)+n_(solvent))`
`=1/(1+55.6)`
`=0.0177`
471.

If equivalent mass of suiphur in `SCl_(2)` is `16 u`, then equivalent mass of `S` in `D_(2)Cl_(2)` will beA. 16B. 64C. 8D. 32

Answer» Correct Answer - 4
In both the compounds equal number of equivalents of `S` and `Cl` combine with each other.
From the given data (atomic mass of `S` is `32 u` and eq mass of S is `16 u`), we conclude that in `SCl_(2)`, 1 mol of `S` atoms `(mass=32 g)` contains 2 equivalent of `S` as the mass of 1 equivalent is `16 g`. We also notice that in `SCl_(2)`
1 mole of `S` atom combines with 2 moles of `Cl` atoms
or
2 equivalents of `S` combines with 1 mole of `Cl` atoms.
or
1 equivalent of `S` combines with 1 mole of `Cl` atoms.
Thus, 1 mole of `Cl` atoms = 1 equivalent of
`Cl=35.5 g`
Now we consider the compound `S_(2)Cl_(2):`
2 moles of `Cl` atoms combine with 2 moles of `S` atoms
or
2 equivalents of `Cl` combine with 2 moles of `S` atoms
or
1 equivalent of `Cl` combines with 1 mole of `S` atoms.
Thus, 1 mole of `S` =1 equivalent of `S=32g`
472.

The equivalent mass of `MnSO_(4)` is half its molecular mass when it is converted toA. `MnO_(4)^(2-)`B. `Mn_(O_(3))`C. `MnO_(2)`D. `MnO_(1)^(-)`

Answer» Correct Answer - 3
`overset(+2)(M)nSO_(4) rarr overset(+6)(M)nO_(4)^(2-)`
`overset(+2)(M)nSO_(4) rarr overset(+3)(M)n_(2)O_(3)`
`overset(+2)(M)nSO_(4) rarr overset(+4)(M)nO_(2)`
`overset(+2)(M)nSO_(4) rarr overset(+6)(M)nO_(4)^(-)`
Only in `3^(rd)` option, change in oxidation number `(ON)` of `Mn` per formula unit of `MnSO_(4)=2`. Thus,
Equivalent mass of `MnSO_(4)=("Formula mass")/2`
473.

What is the fraction of the solute in a `1.00` m aqueous solution ?A. 0.0354B. 0.0177C. 0.177D. 1.77

Answer» Correct Answer - 2
`1.00 m` (one molal) aqueous solution means that 1 mol of the solute is dissolved in every `1 kg (1000 g)` of water. Thus
`n_(solute)=1`
`nH_(2)O=m_(H_(2)O)/(mm_(H_(2)O))=(1000 g)/(18 g mol^(-1))`
`=55.6 mol`
`chi_(solute)=n_(solute)/(n_(solute)+n_(H_(2)O))`
`=(1 mol)/(1 mol+55.6 mol)`
`=0.0177`
474.

The relationship between osmotic pressure at `273 K` when `10 g` glucose `(P_(1)), 10 g` urea `(P_(2))` and `10 g` sucrose `(P_(3))` are dissolved in `250 mL` of water is:A. `P_(1) gt P_(2) gt P_(3)`B. `P_(3) gt P_(1) gt P_(2)`C. `P_(2) gt P_(1) gt P_(3)`D. `P_(2) gt P_(3) gt P_(1)`

Answer» Correct Answer - C
Colligative property is inversely proportional to the molecular mass. Smaller is mol mass of solute, more is osmotic pressure.
475.

`18 g` glucose and `6 g` urea are dissolved in `1 L` aqueous solution at `27^(@)C`. The osmotic pressure of the solution will be a.`8.826 atm`, b.`4.926 atm`,c.`2.92 atm`, d.`4.42 atm`A. 3.826 atmB. 4.926 atmC. 2.92 atmD. 9.42 atm

Answer» Correct Answer - B
`piV=(n_(1)+n_(2))RT`
`piV=((w_(1))/(m_(1))+(w_(2))/(m_(2)))RT`
`pixx1=((18)/(180)+(6)/(60))xx0.0821xx300`
`pi = 4.926` atm]
476.

Two solutions of glucose have osmotic pressure `1.5` and `2.5 atm`,respectively. `1 L` of first solution is mixed with `2 L`of second solution. The osmotic pressure of the resultant solution will be a.`2.62 atm`, b.`6.12 atm`,c.`3.26 atm`, d.`2.16 atm`A. 1.62 atmB. 6.12 atmC. 1.26 atmD. 2.16 atm

Answer» Correct Answer - D
`pi_(1)V_(1)+pi_(2)V_(2)=pi_(R)(V_(1)+V_(2))`
`1.5xx1+2.5xx2=pi_(R)xx3`
`pi_(R)=(6.5)/(3)=2.16` atm]
477.

Statement-1:Isoelectric point is pH at which colloidal can move towards either of electrode Statement-2:At isoelectric point, colloidal solution become electrically neutralA. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-7B. STATEMENT -1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-8C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true

Answer» Correct Answer - D
At isoelectric point, colloid particles will not move towards either of electrode because they will come chargless
478.

Molality of an aqueous solution that produce an elevation of boiling point of 1.00 K at 1 atm pressure. `K_(b)` for water is `0.512 K kg mol^(-1)`A. `0.512m`B. `0.195m`C. `1.95m`D. `5.12m`

Answer» Correct Answer - C
`Delta T_(b)=K_(b)xxm`
`therefore m=(Delta T_(b))/(K_(b))=(1)/(0.512)=1.953 m`
479.

`xg` of `NaCl` (`M=58.5`) put in a dry flask and water was added with continous strring to produce exactly `1L` solution having a molality of `2.00`. Find `x` using the data below: ltbr. Weight `% NaCl` Weight `% NaCl 0 10 12 14` ltbr. Density of solution in `g//mL` 1.0591 1.0742 1.0895 1.1049`

Answer» Correct Answer - `112.82g`
Molality `=("wt of solute//molar mass of solute")/(("wt of solvent in "gm)/(1000))`
`2=((x)/(58.5))/((W)/(1000))`
or `(x)/(W)=(2xx58.5)/(1000)=0.117`
Now we find `%` wt. of `NaCl`
i.e. `(x)/(x+W)xx100=10.4`
So with the help of table given.
Density of solution in `gm//mL=1.0742`
Now for `1` it solution
wt of solution `=1074.2gm`
`therefore` wt of solution `=(1074.2-X)gm=W`
`because (x)/(W)=0.117` or `(x)/(1074.2-x)=0.117`
`therefore x=112.5 gram`.
480.

On the basic intermolecular force predict the correct order of decreasing bolling point of the compound ?A. `CH_(3)OHgtH_(2)gtCH_(4)`B. `CH_(3)OHgtCH_(4)gtH_(2)`C. `CH_(4)gtCH_(3)OHgtH_(2)`D. `H_(2)gtCH_(4)gtCH_(3)OH`

Answer» The order of force attreaction and boiling point is `CH_(3)OHgtCH_(4)gtH_(2)`
481.

Assuming the formation of an ideal solution, determine the boiling point of `a` mixture containing `1560g` benzene (molar mass `=78`) and `1125g` chlorobenzene (molar mass `=112.5`) using the following against an external pressure of `1000` Torr. A. `90^(@)C`B. `100^(@)C`C. `110^(@)C`D. `120^(@)C`

Answer» Correct Answer - B
482.

On mixing `10 mL` of acetone with `40 mL` of chloroform,the total volume of the solution isA. `lt50 mL`B. `gt50 mL`C. `=50 mL`D. Cannot be predicted.

Answer» Correct Answer - A
The interparticle forces in between `CHCl_(3)` and acetone increase due to H-bonding and thus `Delta_("mixing")V` becomes negative.
483.

On dissolving `0.25 g`of a non-volatile substance in`30 mL` benzene `(density 0.8 g mL^(-1))`, its freezing point decreases by `0.25^(@)C` . Calculate the molecular mass of non-volatile substance `(K_(f) = 5.1 K kg mol^(-1))`.

Answer» Mass of benzene, W = volume `xx` density
`=30 xx 0.8 = 24g`
Given, `K_(f)=5.12 "K kg mol"^(-1),w=0.25g,Delta T=0.40^(@)C`
We know that,
`m=(1000K_(f)xxw)/(WxxDeltaT)`
`= (1000xx5.12xx0.25)/(24xx0.40)=133.33`.
484.

On dissolving 0.25 g of a non-volatile substance in 30 mL of benzene (density 0.8 g `mL^(-1)`), its freezing point decreases by `0.40^(@)C`. Calculate the molecular mass of the non-volatile substance. `K_(f)` for benzene is 5.12 `Km^(-1)` .

Answer» Correct Answer - 133.33 g `mol^(-1)`
`W_(B)=0.25g, W_(A)=30xx0.8=24g, 0.024 kg , DeltaT_(f)=0.40^(@)C=0.40 K`
`K_(f)=5.12" K kg mol"^(-1), M_(B)=?`
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((5.12" K kg mol"^(-1))xx(0.25 g))/((0.40 K)xx(0.024 kg ))=133.33" g mol"^(-1)`.
485.

A solution contaning 18 g of non-volatile solute in 200 g of water freezes at 272.07 K . Calculate the molecular mass of solute?

Answer»

∆ Tf = 273-272.07 =0.93

∆ Tf = (Kf x WB )/(MB x WA) x1000

 0.93 = (1.86 x 18 )/ (MB x 200)x1000

MB =  180 g mol-1

486.

A solution containing 18 g of anon-volitile solute in 200 g of waer freezes at 272.07 K. Calculate the molecular mass of the solute. (Given `K_(f)`=1.86 K/m)

Answer» Correct Answer - 180 g `mol^(-1)`
`W_(B)=18g, W_(A)=0.2Kg, K_(f)=1.86," K kg mol"^(-1), T_(f)=273 K, T_(f)=272.07 K`
`DeltaT_(f)=T_(f)^(@)-T_(f)=273-272.07=0.93 K, M_(B)=?`
`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))xx(18g))/((0.93K)xx(0.2 Kg))=180" g mol"^(-1).`
487.

`x g` of non-electrolytic compound (molar mass =200) is dissolved in `1.0 L` of `0.05 M NaCl` solution. The osmotic pressure of this solution is found to be `4.92 atm` at `27^(@)C`. Calculate the value of `x`. Assume complete dissociation of `NaCl` and ideal behaviour of this solution.

Answer» a. For `NaCl : pi =iCRT = 2xx0.05xx0.0821xx30`
`= 2.463` ,
b.For unknown compound:
`pi=CRT=pi/200xx0.0821xx300=0.1231x atm`
Total osmotic pressure `pi=pi_(1)+pi_(2)`
`4.92=2.463+0.1231x`
`x = 19.959 g`
488.

Blood freezes at `272.44 K` and a solution of `3.0 g` of urea in `250 g` of water freezes at `272.63 K`. Calculate the osmotic pressure of blood at `300 K`. (Assume density of blood at `300 K` to be `1 g c c^(-1)`)

Answer» In this question first calculate `K_(f)` of water from urea solution.
`DeltaT_(f)=K_(f)(((W_(B))/(Mw_(B)))/(W_(A))xx1000)`
`rArrK_(f)=(DeltaT_(f))/(((W_(B))/((Mw_(B))/(W_(A)))xx1000))`
`implies K_(f)=(0.37)/((((3)/(60))/(250)xx1000))=1.85`
`[Delta T_(f)=273-272.63=0.37 K]`
Now determine the molality of blood by using
`DeltaT_(f)=K_(f)m`
Now `DeltaT_(f)=273-272.44=0.56^(@)C`
`m=(DeltaT_(f))/(K_(f))=0.56/1.85=0.303`
`rArrMolarity=molality=0.303`
Now using `pi=CRT`
`rArr pi =0.303xx0.082xx300=7.46 atm`
489.

Find the osmotic pressure M/20 solution of urea at `27^(@)C (S = 0.0821 lit atm K^(-1) mol^(-1))`

Answer» Applying the equation `PV = n.ST`,
`P=(n)/(V).ST`
or P = Molarity `xx S xx T`
Given, molarity `=(1)/(2)=0.05,S=0.0821 and T = 27+273=300K`
Substituting values,
`P = 0.05 xx 0.0821 xx 300 = 1.2315` atm.
490.

`0.85%` aqueous solution of `NaNO_(3)` is apparently `90%` dissociated at `27^(@)C`. Calculate its osmotic pressure. `(R = 0.0821 atm K^(-1) mol^(-1)`)

Answer» Correct Answer - 4.67 atm
Solute-B, Water-A
`0.85%("W/A")implies"W"_("B")=0.85"gm, V"_("sol")=100"mL"=(1)/(10)"litre" `
For `"NaNO"_(3),"n=2"`
`"I"=1+("n"-1)alpha`
`"i=1"+alpha`
`"i=1.9"`
`pi="i"xx("w"_("B")"RT")/("m"_("B")xx"V"_("sol")"(L)")`
`=1.9xx(0.85xx0.082xx300)/(85xx0.1)=4.67"atm"`
491.

On a hill station water boils at `98^(@)C`. Amount of salt (NaCl) which should be addedto make its boilings point `100^(@)C` is (`K_(b)=0.52" K kg mole"`)A. 112.5 g/LB. 281.2 g/LC. 225 g/LD. 140.6 g/L

Answer» Correct Answer - 1
492.

When 2.56 g of sulphur is dissolved in 100 g of `CS_(2)`, the freezing point of the solution gets lowerd by 0.383 K. Calculate the formula of sulphur `(S_(x))`. [Given `K_(f)` for `CS_(2)=3.83` "K kg mol"^(-1)`], [Atomic mass of sulphur=32g `mol^(-1)`]

Answer» Calculation of molar mass of sulphur `(S_(x))`
`M_(B)-(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Mass of sulphur `(W_(B))=2.56g`.
Mass of `CS_(2)(W_(A))=100g=0.1 kg`
Molal depression constant `(K_(f))=3.83 K kg mol^(-1)`
Derssion in freezing point `(DeltaT_(f))=0.383 K`
`M_(B)=((3.83 K kg mol^(-1))xx(2.56g))/((0.383K)xx(0.1 kg))=256 mol^(-1)`
Calculate of mlecular fromula of sulphr
Gram atomic mass of sulphur= `32 g mol^(-1)`
In `S_(x),xxx(32 g mol^(-1))=256 g mol^(-1)`
`x=((256 g mol^(-1)))/((32 g mol^(-1)))=8`
Molexular formula of sulphr is `S_(8)`.
493.

The molality of 1M `NaNO_(3)` solution is (d=1.25 g/ml)A. 0.8 mB. 0.858 mC. 1.6 mD. 1 M `Na_(2)SO_(4)`

Answer» Correct Answer - 2
494.

At same temperature, which pair of the following solutions are isotonic solutions?A. `0.2M BaCI_(2)` and `0.2` ureaB. `0.1M` urea and `0.1M NaCI`C. `0.1M NaCI` and `0.1M K_(2)SO_(4)`D. `0.1M Ba (NO_(3))_(2)` and `0.1M Na_(2)SO_(4)`.

Answer» Correct Answer - D
Two solutions are isotonic if they have same molar concentrations of particles in solution.
`0.1 M Ba(NO_(3))_(2)=0.3 M` ion concentration
`0.1 M NaSO_(4) = 0.3 M` ion concentration
495.

At same temperature. which pair of the following solutions are isotonic? (a) 0.2 M BaCl2 and 0.2M urea (b) 0.1 M glucose and 0.2 M urea (c) 0.1 M NaCl and 0.1 M K2SO4 (d) 0.1 M Ba(NO3)2 and 0.1 M Na2SO4

Answer»

(d) 0.1 M Ba(NO3)2 and 0.1 M Na2SO4

0.1 x 3 ion [Ba2 + 2NO3-], 0.1 x 3 ion [2Na+ , SO4-]

496.

2.56 f of suphur in 100 g of `CS_(2)` has depression in freezing point of 0.010^(@) K_(f)=0.1^(@)` (molal)^(-1)`. Hence atomicity of sulphurin the solution isA. 2B. 4C. 6D. 8

Answer» Correct Answer - 4
497.

Isotonic solutions must have the same _____________.(i) solute(ii) density(iii) elevation in boiling point(iv) depression in freezing point

Answer» (ii) density

(iii) elevation in boiling point
498.

If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol of `H_(2)O`, then % ionization NaCl isA. 0.6B. 0.8C. 0.4D. 1

Answer» Correct Answer - 0.04
499.

What is Van’t Hoff factor?

Answer»

In 1880 van‘t Hoff introduced a factor i, known as the van‘t Hofffactor, to account for the extent of dissociation or association. This factor i is defined as the ratio of Normal molar mass to that of the Abnormal molar mass .It is also defined as i = Observed colligative property/Calculated colligative property.

500.

Which of the following pair of solutions can be expected to be isotonic at the same temperature ?A. `0.1 M urea and 0.1 M NaCl `B. 0.1 M NaCl and 0.1 M `Na_(2)SO_(4)`C. 0.1 M `Ca(NO_(3))_(2)` and 0.1 M `Na_(2)SO_(4)`D. 0.1 M glucose and 0.2 M`MgCl_(2)`

Answer» Correct Answer - 3