VITEEE 2019 :: Chemistry :: General questions

• Chemistry - General questions

In the reaction, $C_{6}H_{5}OH\xrightarrow{NaOH}(A)$ $\xrightarrow[{140^0c,(4-7 atm)}]{CO_{2}}$

$(B)\xrightarrow{HCl}  (C)$ the compound (C) is?

Answer:

Explanation:

Treatment of sodium salt of phenol with CO₂ under pressure brings about substitution of the -COOH group for the hydrogen of the ring. This is called as Kolbe's reaction

$A\xrightarrow[{dil.H_{2}SO_{4}}]{K_{2}Cr_{2}O_{7}}$ $B\xrightarrow[{H_{2}O}]{CH_{2}Mgl}$

the reactant A is?

Answer:

Explanation:

$CH_{3}CHOHCH_{3}\xrightarrow[{dil.H_{2}SO_{4}}]{K_{2}Cr_{2}O_{7}}CH_{3}COCH_{3}$ $CH_{3}COCH_{3}\xrightarrow[{H_{2}O}]{CH_{2}Mgl}$

At low pressure, the van der Waal's equation is reduced to

Answer:

Explanation:

When pressure is low

$\left[P+\frac{a}{V^{2}}\right](V-b) = RT$

OR $Z = \frac{PV_{m}}{RT} = 1-\frac{a}{VRT}$

Z = $\frac{-a}{VRT}$

$\frac{PV}{RT}$ =Z

The volume of a closed reaction vessel in which the following equilibrium reaction occurs is halved:

$ 2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)$ As a result

Answer:

Explanation:

$K_{c} = \frac{\left[(SO_3)^2\right]}{\left[(SO_2)^2\right]\left[O_{2}\right]}$

If the volume is reduced to half, K_c will decrease to half. Thus, to maintain the equilibrium, the reaction should shift in the forward direction, i.e. towards the right. Also, to attain equilibrium back, the rate of forward direction will become double the rate of backward direction.

A coordination complex compound of cobalt has a molecular formula containing five ammonia molecules, one nitro group, and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution; on reacting with excess of AgNO₃, AgCl is precipitated. The ionic formula for this complex would be:

Answer:

Explanation:

The most probable complex which gives three mole ions in aqueous solution maybe 

$\left[Co(NH_{3})_{5}NO_{2})\right]Cl_{2}$

• Chemistry - General questions