VITEEE 2017 :: Chemistry :: General questions

• Chemistry - General questions

Knowing that the chemistry of lanthanoids(Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect?

Answer:

Explanation:

Most of the Ln³⁺  compounds except La³⁺ and Lu³⁺ are coloured due to the presence of f-electrons.

An organic compound (A) on reduction gives compound(B). (B) on treatment with CHCl₃, and alcoholic KOH gives (C). (C) on catalytic reduction gives N-methyl aniline. The compound A is

Answer:

Explanation:

Standard entropies of X₂ , Y₂ and XY₃ are 60, 30 and 50 JK^-1 mol^-1 respectively. For the reaction

$\frac{1}{2}X_{2}+ \frac{3}{2}Y_{2}\rightleftharpoons XY_{3} \triangle H = - 30KJ$ to be at equilibrium, the temperature should be

Answer:

Explanation:

ΔS for the reaction,

$\frac{1}{2}X_{2}+ \frac{3}{2}Y_{2}\rightleftharpoons XY_{3}$

ΔS = 50-(30+60)= -40J

For equilibrium ΔG  = 0= ΔH - TΔS

T = ΔH/ΔS = -30000/-40 = 750K

A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives

Answer:

Explanation:

Benzaldehyde and formaldehyde, both do not have α - hydrogen atom, so both will undergo Cannizzaro reaction; here formaldehyde will always be oxidised to formate while the other aldehyde (C₆H₅CHO or any other aldehyde not having α-H,viz- Me₃CCHO) will always be reduced to corresponding alcohol (crossed Cannizzaro reaction)

The half-life period of a first-order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301)

Answer:

Explanation:

For first order reaction, 

$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.93}$

$k = \frac{2.303}{t}log\frac{100}{100-99}$

$\frac{0.693}{6.93}$ = $\frac{2.303}{t}log\frac{100}{100-99}$

$\frac{0.693}{6.93}$ = $\frac{2.303\times2}{t}$

t = 46.06 minutes

• Chemistry - General questions