PYQSide ExamsMy practice

Solutions · Question 28

AIPMT 2015 (Cancelled)MCQEasy📈 Topic in 38% papers+4−1
1

Which of them is not equal to zero for an ideal solution?

  1. Vmix
  2. P = Pobserved - PRaoult
  3. Hmix
  4. Smix

Answer

(4) Smix

Solution

As per NCERT, an ideal solution obeys Raoult's law over the entire range of concentration, and for it mixH = 0 and mixV = 0.

∵ It obeys Raoult's law ⇒ Pobserved = PRaoult ⇒ P = 0

∵ Mixing of two components is a spontaneous process in which randomness increases ⇒ Smix > 0 (not zero).

Answer: (4)

AIPMT 2015 (Cancelled)

Keep practising — Solutions

2

Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?

  1. KCl
  2. C6H12O6
  3. Al2(SO4)3
  4. K2SO4
AIPMT 2014
3

Which condition is not satisfied by an ideal solution?

  1. mixV = 0
  2. mixS = 0
  3. Obeyance to Raoult's Law
  4. mixH = 0
NEET 2013 (Karnataka)
4

Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25 are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be (Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u)

  1. 173.9 mm Hg
  2. 615.0 mm Hg
  3. 90.6 mm Hg
  4. 285.5 mm Hg
AIPMT 2012 (Mains)
5

pA and pB are the vapour pressures of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be

  1. pA + xA (pB - pA )
  2. pA + xA (pA - pB )
  3. pB + xA (pB - pA )
  4. pB + xA (pA - pB )
AIPMT 2012 (Prelims)
6

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86 C/m, the freezing point of the solution will be

  1. -0.18 C
  2. -0.54 C
  3. -0.36 C
  4. -0.24 C
AIPMT 2011 (Mains)
7

200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1)

  1. 51022 g mol-1
  2. 122044 g mol-1
  3. 31011 g mol-1
  4. 61038 g mol-1
AIPMT 2011 (Mains)
8

The freezing point depression constant for water is 1.86 C m-1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by -3.82 C. Calculate the van't Hoff factor for Na2SO4.

  1. 2.05
  2. 2.63
  3. 3.11
  4. 0.381
AIPMT 2011 (Prelims)
9

The van't Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively

  1. less than one and greater than one
  2. less than one and less than one
  3. greater than one and less than one
  4. greater than one and greater than one
AIPMT 2011 (Prelims)
10

An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?

  1. Addition of NaCl
  2. Addition of Na2SO4
  3. Addition of 1.00 molal KI
  4. Addition of water
AIPMT 2010 (Prelims)
11

A solution of sucrose (molar mass = 342 g mol-1) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol-1)

  1. -0.372 C
  2. -0.520 C
  3. +0.372 C
  4. -0.570 C
AIPMT 2010 (Prelims)

All Solutions PYQs →

PDFDownload Solutions PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All NEET Chemistry chapters