JEE 2011 :: General Questions :: Chemistry

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The freezing point (in °C) of solution containing 0.1 g of  $K_{3}[Fe(CN)_{6}]$ (molecular weight 329) in 100 g of water $ (K_{f}=1.86 K kg mol^{-1})$ is 

Answer:

Explanation:

Vant Hoff's factor $ (i)= 4(3K^{+}+[Fe(CN)_{6}]^{3-})$ 

Molality= $\frac{0.1}{329} \times \frac{1000}{100}=\frac{1}{329}$

$ \Rightarrow$    $-\triangle T_{f}=iK_{f}.m=4 \times 1.86 \times \frac{1}{329}=2.3 \times 10^{-2} \Rightarrow T_{f}=-2.3 \times 10^{-2} {^{o}}C$

(As freezing point of water is $0^{0} C$)

The major product of the following reaction is

Answer:

Explanation:

As we know, benzene diazonium salt forms a brilliant coloured dye with β-naphthol, the compound under consideration must be p-toluidine (c) as it is a primary aromatic amine. Primary aromatic amine, on treatment with NaNO₂ in dil. HCl forms the corresponding diazonium chloride salt

The following carbohydrate is

Answer:

Explanation:

Oxidation states of the metal in the minerals haematite and magnetite respectively, are

Answer:

Explanation:

Here, the-OH of the hemiacetal group is equatorial therefore, it is a β-pyranose of an aldohexose

To an evacuated vessel with movable piston under the external pressure of 1 atm, 0.1 mole of He and 1.0 mole of an unknown compound (vapour pressure 0.68 atm at 0 °C) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0°C is close to

Answer:

Explanation:

Haematite is Fe₂O₃, in which the oxidation number of iron is III. Magnetite is Fe₃O₄ which is in fact a mixed oxide (FeO.Fe₂O₃ ) . Hence,iron is present in both ll and lll oxidation state.

• General Questions - Physics
• General Questions - Chemistry
• General Questions - Mathematics