Zinc II Sulphate

Q.

Explain the reason behind the difference in color of the two compounds ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{CuSO}}_4$.

Q. Which is not true about White Vitriol?

1. It can be formed by the reaction of $Zn$ with dilute $H_{2}S{O}_4$
2. It can be formed by the reaction of $ZnO$ with dilute $H_{2}S{O}_4$
3. It can be formed by the reaction of $ZnC{O}_3$ with dilute $H_{2}S{O}_4$
4. When it is heated between $70{}^{\circ }C$ and $280{}^{\circ }C$, anhydrous zinc sulphate is formed

Q. Consider the following reactions (unbalanced)
$Zn+\text{hot conc.}{H}_{2}S{O}_4\to \text{G+R+X}$
$Zn+\text{conc.}NaOH\to \text{T+Q}$
$G+H_{2}S+N{H}_{4}OH\to Z\left(aprecipitate\right)+X+Y$
Choose the correct option(s).

1. The oxidation state of $Zn$ in T is + 1
2. R is a V-shaped molecule
3. Z is dirty white in colour
4. Bond order of Q is 1 in its ground state

Q.

The ——— (physical/chemical) state of the reactants and products are mentioned along with the conditions required for the chemical reaction.

Q. Consider the following reactions (unbalanced)
$Zn+\text{hot conc.}{H}_{2}S{O}_4\to \text{G+R+X}$
$Zn+\text{conc.}NaOH\to \text{T+Q}$
$G+H_{2}S+N{H}_{4}OH\to Z\left(aprecipitate\right)+X+Y$
Choose the correct option(s).

1. The oxidation state of $Zn$ in T is + 1
2. R is a V-shaped molecule
3. Z is dirty white in colour
4. Bond order of Q is 1 in its ground state

Q. The hydrogen phosphate of certain metal has $MHP{O}_4$ formula. The formula of metal chloride is:

1. $MCl$
2. $MC{l}_3$
3. $M_{2}Cl$
4. $MC{l}_2$

Q. In the cyanide process for extraction of gold and silver from ores, the cyanide solution acts as a :

1. Reducing agent to reduce the gold and silver compounds present in the ores into the metallic states
2. Leaching agent to bring the gold and silver into solution as cyanide complexes and thus separate these metals from the ores
3. Leaching agent to bring the ores into solution
4. Leaching agent to dissolve all the other constituents of the ores leaving the gold and silver as metals

Q. Anhydrous calcium chloride is often used as a dessicant. In the presence of excess of $CaC{l}_2$, the amount of the water taken up is governed by $K_p=6.4\times {10}^{8}5$ for the following reaction at room temperature, $CaC{l}_2\left(s\right)+6H_{2}O\left(g\right)\rightleftharpoons CaC{l}_2.6H_{2}O\left(s\right).$ What is the equilibrium vapour pressure of water in a closed vessel that contains $CaC{l}_2\left(s\right)$?

1. $P_{H_{2}O}=6\times {10}^{-15}atm$
2. $P_{H_{2}O}=9\times {10}^{-15}atm$
3. $P_{H_{2}O}=7\times {10}^{-15}atm$
4. $P_{H_{2}O}=5\times {10}^{-15}atm$

Q. (i) A black mineral (A) on heating in presence of air gives a gas (B).
(ii) The mineral (A) on reaction with dilute $H_{2}S{O}_4$ gives a gas (C) and solution of a compound (D).
(iii) On passing gas (C) into an aqueous solution (B), a white turbidity is obtained.
(iv) The aqueous solution of compound (D) on reaction with potassium ferricyanide gives a blue compound (E).

Q. Consider the following reactions (unbalanced)
$Zn+\text{hot conc.}{H}_{2}S{O}_4\to \text{G+R+X}$
$Zn+\text{conc.}NaOH\to \text{T+Q}$
$G+H_{2}S+N{H}_{4}OH\to Z\left(aprecipitate\right)+X+Y$
Choose the correct option(s).

1. The oxidation state of $Zn$ in T is + 1
2. R is a V-shaped molecule
3. Z is dirty white in colour
4. Bond order of Q is 1 in its ground state

Q. Hydrometallurgy is generally used in the case of low grade ores, which cannot be concentrated economically. If true enter 1 else 0.

Q. Write the reaction of $F_2$ with water.

Q. $Pb{O}_2$ is obtained from:

1. the reaction of $Pb$ with air at room temperature
2. the reaction of $PbO$ with $HCl$
3. the reaction of $P{b}_3{O}_4$ with $HN{O}_3$
4. thermal decomposition of $Pb(N{O}_3{)}_2$ at ${200}^{o}C$

Q. Consider the following reactions (unbalanced)
$Zn+\text{hot conc.}{H}_{2}S{O}_4\to \text{G+R+X}$
$Zn+\text{conc.}NaOH\to \text{T+Q}$
$G+H_{2}S+N{H}_{4}OH\to Z\left(aprecipitate\right)+X+Y$
Choose the correct option(s).

1. The oxidation state of $Zn$ in T is + 1
2. R is a V-shaped molecule
3. Z is dirty white in colour
4. Bond order of Q is 1 in its ground state

Q. Sodium thiosulphate is called $\stackrel{I}{.......}$ and is used like $\stackrel{II}{.......}$ and removes unreacted light sensitive $AgBr$ by forming the complex $\stackrel{III}{.......}$

1. $I:hypo,II:fixer,III:\left[Ag\right(S_2{O}_3{)}_2{]}^{3-}$
2. $I:hypo,II:developer,III:[Ag{S}_2{O}_3{]}^{-}$
3. $I:hypo,II:fixer,III:\left[A{g}_2\right(S_2{O}_3\left)\right]$
4. None of the above is correct.

Q. Identify the incorrect reaction (s).

1. $K_{2}C{r}_2{O}_7+4NaCl+3H_{2}S{O}_4(conc.)⟶2Cr{O}_{2}C{l}_2+2NaS{O}_4+K_{2}S{O}_4+3H_{2}O$
2. $K_{2}C{r}_2{O}_7+6Kl+7H_{2}S{O}_4(conc.)⟶3l_2+C{r}_2\left(S{O}_4\right)+4K_{2}S{O}_4+7H_{2}O$
3. $K_{2}C{r}_2{O}_7+4AgCl+3H_{2}S{O}_4(conc.)⟶2Cr{O}_{2}C{l}_2+2A{g}_{2}S{O}_4+K_{2}S{O}_4+3H_{2}O$
4. $Mn{O}_2+NaCl+2H_{2}S{O}_4(conc.)⟶NaHS{O}_4+MnS{O}_4+HCl+H_{2}O+1/2O_2$

Q. $ZnS{O}_4.7H_{2}O\underset{Below{70}^{\circ }C}{\overset{Above{39}^{\circ }C}{\to }}A\stackrel{Above{70}^{\circ }C}{\to }B\stackrel{Above{280}^{\circ }C}{\to }{O}_2+S{O}_2+ZnO$. A and B respectively are:

1. $ZnS{O}_4.H_{2}O\&ZnS{O}_4.6H_{2}O$
2. $ZnS{O}_4.2H_{2}O\&ZnS{O}_4.6H_{2}O$
3. $ZnS{O}_4.6H_{2}O\&ZnS{O}_4.2H_{2}O$
4. $ZnS{O}_4.6H_{2}O\&ZnS{O}_4.H_{2}O$

Q. $NaOH$ can be used to separate ${Al\left(OH\right)}_3$ and ${Zn\left(OH\right)}_2$.

1. yes
   
2. no
   
3. in high pressure only
   
4. none of the above
5. in High temperature only

Q. in the reaction $C_2{H}_5{OH}_2{H}_5+CO\begin{array}{c}{BF}_{3500atms}\\ ⟶\\ {150}^{\circ }C\end{array}X$. What is $X$?

1. $Diethylcarbonate$
2. $Ethylcarbonate$
3. $Diethylperoxide$
4. $Ethylpropionate$

Q. The product formed in contact tower is :

1. $S{O}_3$
2. $S{O}_2$
3. $H_2{S}_2{O}_8$
4. $H_2{S}_2{O}_7$

Q. $ZnS{O}_4.7H_{2}O\underset{Below{70}^{\circ }C}{\overset{Above{39}^{\circ }C}{\to }}A\stackrel{Above{70}^{\circ }C}{\to }B\stackrel{Above{280}^{\circ }C}{\to }{O}_2+S{O}_2+ZnO$. A and B respectively are:

1. $ZnS{O}_4.H_{2}O\&ZnS{O}_4.6H_{2}O$
2. $ZnS{O}_4.2H_{2}O\&ZnS{O}_4.6H_{2}O$
3. $ZnS{O}_4.6H_{2}O\&ZnS{O}_4.2H_{2}O$
4. $ZnS{O}_4.6H_{2}O\&ZnS{O}_4.H_{2}O$

Q. White vitriol is:

1. $ZnS$
2. $ZnS{O}_4$
3. $ZnS{O}_4.7H_{2}O$
4. $ZnC{O}_3$