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 Multiple Choice QuestionsMultiple Choice Questions

181.

50 cm3 of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration was discontinued after adding 50 cm3 of NaOH. The remaining titration is completed by adding 0.5 N KOH. The volume of KOH required for completing the titration is 

  • 12 cm3

  • 10 cm3

  • 21.0 cm3

  • 16.2 cm3


182.

The rms velocity of hydrogen is 7 times the rms velocity of nitrogen. If T is the temperature of the gas, which of the following is true?

  • TH2 = 7  TN2

  • TN2 = TH2

  • TN2 = 7 TH2

  • TN2 = 2TH2


183.

50 cm3 of 0.2 NHCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding 50 cm3 of NaOH.The remaining titration is completed by adding 0.5 N KOH. The volume of KOH required for completing the titration is

  • 12 cm3

  • 10 cm3

  • 25 cm3

  • 10.5 cm3


184.

1 g of silver gets distributed between 10 cm3 of molten zinc and 100 cm3 of molten lead at 800°C. The percentage of silver in the zinc layer is approximately

  • 89

  • 91

  • 97

  • 94


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185.

Excess of silver nitrate solution is added to 100 mL of 0.01 M pentaaqua chloro chromium (III) chloride solution. The mass of silver chloride obtained in grams is [Atomic mass of silver is 108].

  • 287 × 10-3

  • 143.5 × 10-3

  • 143.5 × 10-2

  • 287 × 10-2


186.

The empirical formula of a non-electrolyte is CH2O. A solution containing 3 g of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The
molecular formula of the compound is 

  • CH2O

  • C2H4O2

  • C4H8O4

  • C3H6O3


187.

The empirical formula of a non-electrolyte is CH2O. A solution containing 6g of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution at the same temperature. The molecular formula of

  • C2H4O2

  • C3H6O3

  • C5H10O5

  • C4H8O4


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188.

The total number of electrons in 18 mL of water (density = 1 g mL-1) is

  • 6.02 × 1025

  • 6.02 × 1024

  • 6.02 ×  18 × 1023

  • 6.02 × 1023


B.

6.02 × 1024

In 18 mL, number of moles of

H2O = mass of H2Omolecular mass

= density × volumemolecular mass= 1 × 1818

= 1 mol

 Number of moles of H2O in 1 mol = 6.022 × 1023

and number of e- in 1 molecule of H2O = 1 × 2 + 8 = 10

 Number of e- in 1 molecule of H2

= 6.022 × 1023 × 10

= 6.022 × 1024


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189.

The number of water molecules present in a drop of water weighing 0.018 g is

  • 6.022 × 1026

  • 6.022 × 1023

  • 6.022 × 1019

  • 6.022 × 1020


190.

Empirical formula of a compound is CH2O and its molecular mass is 90, the molecular formula of the compound is

  • C3H6O3

  • C2H4O2

  • C6H12O6

  • CH2O


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