Chemistry MCQs for NEET — Practice Questions with Answers

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The relationship between osmotic pressure at 273K, when 10gm glucose(P1), 10 gm urea (P2) and 10 gm sucrose (P3) dissolved in 250 ml of water is

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Explanation

p=iMRT

i=dimensionless van't Hoff Factor 

M is the molarity

R=0.0826 L.atm mol-1K-1

T is the temperature

Sucrose = 342

Urea = 60

Glucose = 180

P2>P1>P3 

If 'A' contains 2% NaCl and is separated by a semi permeable membrane from 'B', which contains 10% NaCl, which event will occur ?

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Explanation

2% NaCl solution (A) seperated by semipermeable membrane from 10% NaCl solution (B). Then from dilute solution to conc. solution solvent flow.

The vapour pressure of water depends upon:

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Explanation

The vapour pressure of water depends upon temperature only.

The boiling point of an azeotropic mixture of water and ethanol is less than that of water and ethanol. The mixture shows

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Explanation

Mixture of ethanol and water shows positive +ve deviation from Roult's law.

Mole of K2SO4 to be dissolved in 12 moles water to lower its vapour pressure by 10 mm of Hg at a temperature at which vapour pressure of pure water is 50 mm of Hg is

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Explanation

 

 

P-PsP=inin+N1050=3×n3×n+123n+12 = 15nn=1

Which has highest freezing point ?

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Explanation

Tf  is directly proportional to  number of ions

Lower is the effective molarity, higher is the freezing point.

Aluminium phosphate is 100% ionized in 0.01m aqueous solution. Hence, Tb/kb is

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Explanation

For AlPO4, i=2

TbKb=i × m =2×0.01=0.02

Which solution has highest freezing point?

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Explanation

Higher is the molar mass, lesser is the effective molarity. So, 1% sucrose solution has highest freezing point.

Osmotic pressure of 40% (w/V) urea solution is 1.64 atm. and that of 3.42% (w/V) cane sugar solution is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution will be?  

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Explanation

π= π1+π2 , if equal volume are mixed volume of solution become double.

20g of naphthoic acid (C11H8O2) dissolved in 50g of benzene (Kf=1.72 K Kg mol-1) shows a depression in freezing point of 2K. The Vant Hoff factor is?

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Explanation

Mol. wt. of naphthoic acid=172

ΔTcal=1000 x 1.72 x 20 /50 x 172 = 4

i=experimentalΔTf=2/4=0.5
    calculated  ΔTf

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