Solid NaHCO3 will be neutralised by 40.0 mL of 0.1 M H2SO4 soluti

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 Multiple Choice QuestionsMultiple Choice Questions

231.

The normality of 10% (w/v) of acetic acid is

  • 1 N

  • 1.3 N

  • 1.7 N

  • 1.9 N


232.

Consider the following solutions, 

A = 0.1 M glucose,

B = 0.05 M NaCl,

C = 0.05 M BaCl2

D = 0.1 M AlCl3

Which of the following pairs is isotonic?

  • A and B

  • A and D

  • A and C

  • B and C


233.

Which of the following aqueous solution should have a highest boiling point?

  • 1.0 M NaOH

  • 1.0 M Na2SO4

  • 1.0 M NH4NO3

  • 10 MKNO3


234.

Combustion of glucose takes place according to the equation.

C6H12O6 + 6O2 → 6CO2 + 6H2O

Δ = - 72 k-cal

The energy required for combustion of 1.6 g of glucose is

  • 0.064 kcal

  • 0.64 kCal

  • 6.4 kcal 

  • 64 kcal


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235.

when the heat of reaction at constant pressure is -2.5 x 103 cal can entropy change is 7.4 cal deg-1 at 25oC, the reaction is predicted as

  • reversible

  • Spontaneous

  • non -spontaneous

  • Irreversible


236.

The volume of water to be added to 100 cm3 of 0.5 NH2SO4 to get decinormal concentration is

  • 100 cm3

  • 450 cm3

  • 500 cm3

  • 400 cm3


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237.

Solid NaHCO3 will be neutralised by 40.0 mL of 0.1 M H2SO4 solution. What would be the weight of solid NaHCO3 in gram?

  • 0.672 g

  • 6.07 g

  • 17 g

  • 20 g


A.

0.672 g

The required neutralisation reaction is

2NaHCO+ H2SO4 Na2SO+ 2H2O + 2CO2

Number of moles in 2NaHCO3 and Na2SO4 are 2 and 1 respectively.

Therefore, 2NaHCO3 = 168 gm and Na2SO4 = 98 gm

Moles of H2SO4 = M × VmL = 40.0 ×0.1 = 4m- mole. Also, it can written as m- moles of NaHCO3 when neutralised = 4 × 2 = 8m-moles.

m-mole = wm×1000

 8 = w85×1000

w = 84 × 81000 = 0.672 gms.


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238.

A solution with negative deviation among the following is

  • ethanol-acetone

  • chlorobenzene-bromobenzene

  • chloroform-acetone

  • benzene-toluene


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239.

For an ideal binary liquid solution with px0 > py0 in which relation between Xx (mole fraction of X in liquid phase) and Yx ( mole fraction of Y in liquid and vapour phase respectively.

  • Xx > YX

  • Xx = Yx

  • XXXY < YXYY

  • XX, YX, XY and YY cannot be correlated


240.

In an experiment, addition of 4.0 mL of 0.005 M BaCl2 to 16.0 mL of arsenious sulphide sol just causes complete coagulation in 2 h. The flocculating value of the effective ion is

  • Ba2+, 1.0

  • Ba2+, 2.0

  • Cl-, 1.0

  • Cl-, 2.0


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