Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

451.

0.6 g of a solute is dissolved in 0.1 littre of a solvent which develops an osmotic pressure of 1.23 atm at `27^(@)C`. The molecular mass of the substance isA. `149.5 "g mole"^(-1)`B. `120 "g mole"^(-1)`C. `430 "g mole"^(-1)`D. None of these

Answer» Correct Answer - B
`m=(wRT)/(PV)=(0.6xx0.082xx300)/(1.23xx0.1)=120`
Discussion
452.

For getting accurate value of molar mass of a solute by osmotic pressure measurementA. The solute must be volatileB. The solution concentration must be highC. The solute should undergo dissociationD. The solute must be non-volatile

Answer» Correct Answer - D
To get precise values of molar mass of a solute by osmotic pressure measurement, the solute should be non-volatile.
Discussion
453.

The molar mass of the solute sodium hydrdoxide obtained from the measurement of the osmotic pressure of its aqueous solution at `27^(@)C` is `25 g mol^(-1)`. Therefore its ionization percentage in this solution isA. 75B. 60C. 80D. 70

Answer» Correct Answer - B
In case of ionisation of binary electrolyte
Abnormal molecular mass `=(M_("normal"))/(1+alpha)`
Abnormal molecular mass = 40
So, `25=(40)/(1+alpha)rArr alpha =0.6`
Ionisation percemntage `= 0.6xx100=60%`.
Discussion
454.

The osmotic pressure of one molar solution at `27^(@)C` isA. 2.46 atmB. 24.6 atmC. 1.21 atmD. 12.1 atm

Answer» Correct Answer - b
`p = (n)/(V)RT = MRT` [ Molarity (M) `= (n)/(V)`]
`= 1xx0.0821xx300 = 24.6 atm`
Discussion
455.

A substance will be deliquescent if its vapour pressure is:A. is equal to the atmospheric pressureB. is equal to that of water vapour in airC. is less than that of water vapour in airD. is greater than that of water vapour in air

Answer» Correct Answer - c
Discussion
456.

The boiling point of an aqueous solution of a non-volatile solute is `100.15^(@)C`. What is the freezing point of an aqueous solution obtained by dilute the above solution with an equal volume of water. The values of `K_(b)` and `K_(f)` for water are `0.512` and `1.86^(@)C mol^(-1)`:A. `-0.544^(@)C`B. `-0.512^(@)C`C. `-0.272 ^(@)C`D. `-1.86 ^(@)C`

Answer» Correct Answer - C
For same solution `(Delta T_(f))/(Delta T_(b))=K_(f)/K_(b) ` or `Delta T_(f) = Delta T_(b) xx K_(f)/K_(b)`
or `Delta T_(f) = (0.15 xx 1.86)/20.512=0.545`
Now on diluting the solution to double
` :. Delta T_(f) = 0.545/2=0.272`
`:. f. pt. = -0.272^(@) C`
Discussion
457.

A solution containing 3.3 g of a substance in 125g of benzene `(b.p 80^(@)C)` boils at `80.66^(@)C.` If `K_(b)` for one litre of benzene is `3.28^(@)C,` the molecular mass of the substance shall beA. 127.2B. 131.2C. 137.12D. 142.72

Answer» Correct Answer - B
`Delta T_(b) = (K_(b) xx W_(2))/(M_(2) xx W_(1))=1000` `:. M_(2) = (1000K_(b) W_(2))/(W_(1) Delta T_(b))=(1000 xx 3.28 xx 3.3)/(125 xx 0.66)=131.2`
Discussion
458.

2.0 molar solution is obtained, when 0.5 mole solute is dissolved inA. 250 ml solventB. 250 g solventC. 250 ml solutionD. 1000 ml solvent

Answer» Correct Answer - D
We know that
Molarity = `("Number of moles of solute")/("Volume of solution in litre")`
`therefore 2.0 =(0.5)/("Volume of solution in litre")`
`therefore` Volume of solution in litre
`=(0.5)/(2.0)=0.250` litre = 250 ml.
Discussion
459.

171 g of cane sugar `(C_(12) H_(22) O_(11))` is dissolved in 1 litre of water. The molarity of the solution isA. 2.0 MB. 1.0 MC. 0.5 MD. 0.25 M

Answer» Correct Answer - C
Molarity `=(w)/(m.wt. xx "volume in litre")=(171)/(342xx1)=0.5 M`.
Discussion