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A is a binary compound of a univalent me...

A is a binary compound of a univalent metal. `1.422 g` of A reacts completely with `0.321 g` of sulphur in an evacuated and sealed tube to give `1.743 g` of a white crystalline solid B, that forms a bydrated double salt, C with `Al_(2) (SO_(4))_(3)`. Identify A, B and C.

Text Solution

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Compound B forms hydrated crystals with `Al_(2)(SO_(4))_(3)`. Also, B is formed with univalent metal on heating with sulphur. Hence, compound B must has the molecular formula `M_(2)SO_(4)` and compound A must be an oxide of M which reacts with sulphur to give metal sulphate as
`A+S rarr M_(2)SO_(4)`
`:' 0.321 g` sulphur gives `1.743 g` of `M_(2)SO_(4)`
`:. 32.1 g S` (one mole) will give `174.3 g M_(2)SO_(4)`
Therefore, molar mass of `M_(2)SO_(4)=174.3 g`
`implies 174.3=2xx` Atomic weight of `M+32.1+64`
`implies` Atomic weight of `M=39`, metal is potassium `(K)`
`K_(2)SO_(4)` on treatement with aqueous `Al_(2)(SO_(4))_(3)` gives potash-alum.
`underset(B)(K_(2)SO_(4))+Al_(2)(SO_(4))_(3)+24H_(2)O rarr K_(2)SO_(4)Al_(2)underset(C)((SO_(4))_(3)).24H_(2)O`
If the metal oxide A has molecular formula `MO_(x)` two moles of it combine with one mole of sulphur to give one mole of metal sulphate as
`2KO_(x)+S rarr K_(2)SO_(4)`
`implies x=2`, i.e. `A` is `KO_(2)`.
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A is a binary compound of a univalent metal. When 1.422 g of A reacts completely with 0.321 g of sulphur in an evacuated and sealed tube, 1.743 g of white crystalline solid B produced, which produces a hydrated double salt C with Al_(2)(SO_(4))_(3) . Identfy A, B and C.

The metallic salt (XY) is soluble in water. (a) When the aqueous soluble of (XY) is treated with NaOH solution, a white precipitate (A) is formed. In excess of NaOH solution, a white precipitate (A) is formed. In excess of NaOH solution, white precipitate (A) dissolves to form a compound (B) . When this solution is boiled with soild NH_(4) Cl , a precipitate of compound (C) is formed. (b) An aqueous solution on treatment with BaCl_(2) solution gives a white precipitate (D) white is insoluble in conc HCl . ( c) The metallic salt (XY) forms a double salt (E) with potassium sulphate. Identify (XY),(A),(B),(C),(D) and (E) .

An organic compound (A) C_(4)H_(9)Cl on reacting with aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C ), which is also formed on passing the vapours of (B) over the heated copper. The compound (C ) readily decolourises bromine water. Ozonolysis of (C ) gives two compounds (D) and (E ). Compound (D) reacts with NH_(2)OH to give (F) and compound (E ) reacts with NaOH to have an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared form propyne on treatment with water in the presence of Hg^(2+) and H_(2)SO_(4) . Identify (A) to (H) with proper reasoning.

2 g of a metal in H_(2) SO_(4) gives 4.51 g of the metal sulphate. The specific heat of metal is 0.057 cal g^(-1) . Calculate the valency and atomic weight of metal.

Knowledge Check

  • In compound A,1 g of nitrogen combines with 0.57 g oxygen. In compound B,2 g of nitrogen combines with 2.24 g oxygen. In compound C,3 g of nitrogen combines with 5.11 g oxygen. These results obey following law

    A
    Law of constant proportion
    B
    Law of multiple proportion
    C
    Law of reciprocal proportion
    D
    Dalton's law of partial pressure

  • If 50 ml of A_(2)B_(3) reacts completely with 200 ml of C_(2) in closed vessel according to the equation , 2A_(2)B_(3)(g)+5C_(2)(g) to 3C_(3)B_2(g)+CA_(4)(g) . The composition of the mixture of gases is :

    A
    10 ml ` C_(2)` , 25 ml `C_(3)B_(2)`, 100 ml `CA_(4)`
    B
    25 ml `C_(2)`, 75 ml ` C_(3),B_2,` 25 ml ` CA_(4)`
    C
    75 ml ` C_(2)`, 75 ml ` C_(3)B_(2)`, 25 ml ` CA_(4)`
    D
    100 ml `C_(2)`, 50 ml `C_(3)B_(2)`, 50 ml `CA_(4)`

  • An organic compound A reacts with sodium metal and forms B. On heating with conc. H_(2)SO_(4) at 140^(@)C A gives diethyl ether. A and B are respectively

    A
    `C_(2)H_(5)OH` and `C_(2)H_(5)ONa`
    B
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    C
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    D
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