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ICSE Class 10 Chemistry Solved Question Paper 2004

Short Answer Type

51.

Write balanced equations for the following reactions:

Chlorine is passed into an aqueous solution of sulphur dioxide.

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52.

Write balanced equations for the following reactions:

Aluminium powder is warmed with hot and concentrated caustic soda solution.

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53.

Write balanced equations for the following reactions:

Concentrated nitric acid is added to copper turnings kept in a beaker.

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54.

Write balanced equations for the following reactions:

Red lead (trilead tetroxide) is warmed with concentrated hydrochloric acid.

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55.

Write balanced equations for the following reactions:

Chlorine gas is passed through an aqueous solution of Iron (II) sulphate acidified with dilute sulphuric acid.

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56.

Write balanced equations for the following reactions:

Ethane is burnt in air.

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Long Answer Type

57.

i) Name the catalyst which helps in the conversation of sulphur dioxide to sulphur trioxide in step C.

ii) In the contact process for the manufacture of sulphuric acid, sulphur trioxide in not converted to sulphuric acid by the reaction it with water. Instead of a two -steps procedure is used. Write the equation for the two-step involved in D.

iii) What type of substance for the reaction by which sulphur dioxide is converted to sodium sulphite in step E.

iv) Write the equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step E.

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58.
When heated, potassium permanganate decomposes according to the following equation:

$bold 2 bold KMnO subscript bold 4 bold space bold rightwards arrow with bold space bold space bold space bold space bold space bold space on top bold space stack bold left curly bracket bold K subscript bold 2 bold MnO subscript bold 4 bold space bold plus bold MnO subscript bold 2 bold space bold right curly bracket with bold solid bold space bold residue below bold plus bold O subscript bold 2$

i) some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825g, calculate the relative molecular mass of oxygen.

ii) Given that the molecular mass of potassium permanganate is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? ( Molar volume at room temperature is 24 litres.)

i) Loss of weight of due to formation of O₂
the weight of 1 litre of oxygen under room conditions = 1.32g
the weight of 1 litre of hydrogen under room conditions = 0.0825 g

Now according to Avogadro's law under similar conditions of temperature and pressure equal volumes of gases contain equal number of moles

Suppose, 1 litre of oxygen under room condition contain n molecules.

Therefore,
Relative molecular mass of oxygen =

$fraction numerator weight space of space 1 space molecular space of space straight O subscript 2 over denominator weight space of space 1 space atom space of space hydrogen space end fraction equals space fraction numerator weight space of space straight n space molecule space of space straight O subscript 2 over denominator weight space of space 1 divided by 2 space molecule space of space hydrogen end fraction$
[therefore, atom of hydrogen = 1/2 molecules of hydrogen]
= $fraction numerator 1.32 over denominator 0.0825 divided by 2 end fraction space equals 32$

ii) $stack 2 KMnO subscript 4 space with 2 space straight x space 158 space straight g below rightwards arrow stack left curly bracket straight K subscript 2 MnO subscript 4 space plus MnO subscript 2 with solid space residue below right curly bracket space plus space stack straight O subscript 2 with 24 space litre below therefore comma 15.8 straight g space of space KMnO subscript 4 space on space heating space gives space straight O subscript 2 space equals fraction numerator 24 over denominator 2 space straight x space 158 space end fraction straight x space 15.8 space equals 1.2 space straight g$

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59.

Sodium hydroxide solution is added first in a small quantity, then in excess to the aqueous salt solution of copper (II) sulphate,Zinc, lead nitrate, Calcium chloride and iron(III) sulphates. Copy the following table and write the colour of the precipitate in (i) to (v) and the nature of the precipitate (soluble or insoluble ) in (vi) to (x).

Aqueous salt solution	Colour of precipitate when NaOH is added in a small quantity	Nature of precipitate (soluble or insoluble) when NaOH is added in excess.
Copper (II) sulphate	(i)	(vi)
Zinc nitrate	(ii)	(vii)
Lead nitrate	(iii)	(Viii)
Calcium chloride	(iv)	(ix)
Iron (III) sulphate	(v)	(x)

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60.

Which of the following methods,A,B,C D or E is generally used for preparing the chlorides listed below from (i) to (v). Answer by writing down the chloride and the letter pertaining to the corresponding method. Each letter is to be used only once.

A. Action of an acid on a metal.
B. Action of an acid on an oxide or carbonate
C. Direct combination
D. Neutralization of an alkali by an acid.
E. Precipitation (double decomposition)

i) Copper (II chloride)
ii) Iron (II) chloride
iii) Iron (III) chloride
iv) Lead (II) chloride
v) Sodium chloride.

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