500 mL of 1.2 M KI solution is mixed with 500 mL of $0.2 \mathrm{M} \mathrm{KMnO}_4$ solution in basic medium. The liberated iodine was titrated with standard $0.1 \mathrm{M} \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ solution in the presence of starch indicator till the blue color disappeared. The volume (in L ) of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ consumed is $\_\_\_\_$ . (Nearest integer)
Explanation:
In basic medium, $\mathrm{KMnO_4}$ is reduced to $\mathrm{MnO_2}$ and $\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$.
Step 1: Balance the redox reaction in basic medium
Half-reactions:
Reduction:
$ \mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-} $
Oxidation:
$ \mathrm{2I^- \rightarrow I_2 + 2e^-} $
LCM of electrons $=6$:
$ 2(\mathrm{MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-}) $
$ 3(\mathrm{2I^- \rightarrow I_2 + 2e^-}) $
Adding:
$ \mathrm{2MnO_4^- + 4H_2O + 6I^- \rightarrow 2MnO_2 + 8OH^- + 3I_2} $
So,
- $2$ mol $\mathrm{MnO_4^-}$ produce $3$ mol $\mathrm{I_2}$.
Step 2: Calculate moles of reactants
Moles of KI:
$ n(\mathrm{I^-})=0.5\times 1.2=0.6\ \text{mol} $
Moles of $\mathrm{KMnO_4}$:
$ n(\mathrm{MnO_4^-})=0.5\times 0.2=0.1\ \text{mol} $
From the balanced equation:
$ 1\ \text{mol }\mathrm{MnO_4^-} \text{ requires } 3\ \text{mol }\mathrm{I^-} $
So $0.1$ mol $\mathrm{MnO_4^-}$ requires $0.3$ mol $\mathrm{I^-}$, which is available ($0.6$ mol).
Hence, $\mathrm{KMnO_4}$ is the limiting reagent.
Step 3: Moles of $\mathrm{I_2}$ liberated
$ n(\mathrm{I_2})=\frac{3}{2}\times 0.1=0.15\ \text{mol} $
Step 4: Titration with thiosulfate
Reaction:
$ \mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-}} $
So,
$ n(\mathrm{S_2O_3^{2-}})=2\times 0.15=0.30\ \text{mol} $
Given $\mathrm{Na_2S_2O_3}$ is $0.1\ \mathrm{M}$:
$ V=\frac{n}{M}=\frac{0.30}{0.1}=3.0\ \text{L} $
Answer (nearest integer)
$ \boxed{3} $
X and Y are the number of electrons involved, respectively during the oxidation of $\mathrm{I}^{-}$to $\mathrm{I}_2$ and $\mathrm{S}^{2-}$ to S by acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$. The value of $\mathrm{X}+\mathrm{Y}$ is $\_\_\_\_$ .
Explanation:
$ \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O} $
In this reaction, $6 \mathrm{I}^{-}$ ions get oxidised to $3 \mathrm{I}_2$.
Each $\mathrm{I}^{-}$ loses $1$ electron during oxidation, so total electrons lost (in moles) $=\mathrm{X}=6$.
$ \mathrm{Cr}_2 \mathrm{O}_7^{2-}+3 \mathrm{~S}^{2-}+14 \mathrm{H}^{+} \rightarrow+\mathrm{S}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} $
Here, $3 \mathrm{S}^{2-}$ ions get oxidised to $3\mathrm{S}$.
Each $\mathrm{S}^{2-}$ loses $2$ electrons to become $\mathrm{S}$, so total electrons lost (in moles) $=\mathrm{Y}=3 \times 2=6$.
Therefore, $x+y=6+6=12$.
200 cc of $x \times 10^{-3} \mathrm{M}$ potassium dichromate is required to oxidise 750 cc of 0.6 M Mohr's salt solution in acidic medium.
Here $x=$ $\_\_\_\_$ .
Explanation:
In acidic medium, dichromate oxidises $Fe^{2+}$ to $Fe^{3+}$.
Half-reactions (NCERT method):
$Cr_2O_7^{2-}+14H^+ + 6e^- \rightarrow 2Cr^{3+}+7H_2O$
$Fe^{2+}\rightarrow Fe^{3+}+e^-$
So, $1$ mole of $Cr_2O_7^{2-}$ oxidises $6$ moles of $Fe^{2+}$.
Step 1: Moles of $Fe^{2+}$ in Mohr’s salt solution
Volume $=750\ \text{cc}=0.75\ \text{L}$, Molarity $=0.6\ \text{M}$
$n(Fe^{2+}) = 0.75 \times 0.6 = 0.45\ \text{mol}$
Step 2: Moles of $K_2Cr_2O_7$ required
$n(K_2Cr_2O_7)=\frac{0.45}{6}=0.075\ \text{mol}$
Step 3: Use given dichromate solution data
Volume $=200\ \text{cc}=0.2\ \text{L}$, Molarity $=x\times 10^{-3}\ \text{M}$
$n = 0.2 \times x\times 10^{-3} = 2\times 10^{-4}x$
Equate moles:
$2\times 10^{-4}x = 0.075$
$x=\frac{0.075}{2\times 10^{-4}}=375$
$ \boxed{x = 375} $
$\mathrm{KMnO}_4$ acts as an oxidising agent in acidic medium. ' X ' is the difference between the oxidation states of Mn in reactant and product. ' Y ' is the number of ' d ' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of $\mathrm{X}+\mathrm{Y}$ is __________.
Explanation:
X is difference in oxidation state.
$7-2=5$
So $X=5$
$\begin{aligned} & 6 \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O} \\ & \rightarrow\left[\mathrm{Fe}_3\left(\mathrm{OH}_2\right)\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+2 \mathrm{H}^{\oplus} \\ & {\left[\mathrm{Fe}_3(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+4 \mathrm{H}_2 \mathrm{O}} \\ & \rightarrow \underset{\text { Brown red ppt }}{\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right]\right.}+\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}^{\oplus} \end{aligned}$
$\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0$ contains 5 d electrons
So $Y=5$
$X+Y=5+5=10$
Some $\mathrm{CO}_2$ gas was kept in a sealed container at a pressure of 1 atm and at 273 K . This entire amount of $\mathrm{CO}_2$ gas was later passed through an aqueous solution of $\mathrm{Ca}(\mathrm{OH})_2$. The excess unreacted $\mathrm{Ca}(\mathrm{OH})_2$ was later neutralized with 0.1 M of 40 mL HCl . If the volume of the sealed container of $\mathrm{CO}_2$ was $x$, then $x$ is ____________ $\mathrm{cm}^3$ (nearest integer). [Given : The entire amount of $\mathrm{CO}_2(\mathrm{~g})$ reacted with exactly half the initial amount of $\mathrm{Ca}(\mathrm{OH})_2$ present in the aqueous solution.]
Explanation:
Let moles of $\mathrm{CO}_2=\mathrm{n}$
moles of $\mathrm{Ca}(\mathrm{OH})_2$
total initially $=2 \mathrm{n}$
excess $\mathrm{Ca}(\mathrm{OH})_2=\mathrm{n}$
gm equivalent of $\mathrm{Ca}(\mathrm{OH})_2=$ gm equivalent of HCl
$ \begin{aligned} & \mathrm{n} \times 2=0.1 \times \frac{40}{1000} \times 1 \\ & \mathrm{n}=2 \times 10^{-3} \end{aligned} $
Volume of $\mathrm{CO}_2=2 \times 10^{-3} \times 22400=44.8 \mathrm{~cm}^3$
Only $2 \mathrm{~mL}$ of $\mathrm{KMnO}_4$ solution of unknown molarity is required to reach the end point of a titration of $20 \mathrm{~mL}$ of oxalic acid $(2 \mathrm{M})$ in acidic medium. The molarity of $\mathrm{KMnO}_4$ solution should be ________ M.
Explanation:
$\begin{aligned} & \mathrm{(M) \times(2) \times(5)=2 \times 20 \times 2} \\ & \mathrm{M=8} \end{aligned} $
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is $(+)$ _________.
Explanation:
$\begin{aligned} & \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7(\mathrm{~s})+4 \mathrm{KCl}(\mathrm{s})+6 \mathrm{H}_2 \mathrm{SO}_4 \text { (conc.) } \\ & \rightarrow 2 \mathrm{CrO}_2 \mathrm{Cl}_2(\mathrm{~g})+6 \mathrm{KHSO}_4+3 \mathrm{H}_2 \mathrm{O} \end{aligned}$
This reaction is called chromyl chloride test.
Here oxidation state of $\mathrm{Cr}$ is +6.
Number of moles of $\mathrm{H}^{+}$ ions required by $1 \mathrm{~mole}$ of $\mathrm{MnO}_4^{-}$ to oxidise oxalate ion to $\mathrm{CO}_2$ is _________.
Explanation:
$2 \mathrm{MnO}_4^{-}+5 \mathrm{C}_2 \mathrm{O}_4^{2-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O}$
$\therefore$ Number of moles of $\mathrm{H}^{+}$ ions required by 1 mole of $\mathrm{MnO}_4^{-}$ to oxidise oxalate ion to $\mathrm{CO}_2$ is 8
Total number of species from the following which can undergo disproportionation reaction is ________.
$\mathrm{H}_2 \mathrm{O}_2, \mathrm{ClO}_3^{-}, \mathrm{P}_4, \mathrm{Cl}_2, \mathrm{Ag}, \mathrm{Cu}^{+1}, \mathrm{~F}_2, \mathrm{NO}_2, \mathrm{K}^{+}$
Explanation:
Intermediate oxidation state of element can undergo disproportionation.
$\mathrm{H}_2 \mathrm{O}_2, \mathrm{ClO}_3^{-}, \mathrm{P}_4, \mathrm{Cl}_2, \mathrm{Cu}^{+1}, \mathrm{NO}_2$
$2 \mathrm{MnO}_4^{-}+\mathrm{bI}^{-}+\mathrm{cH}_2 \mathrm{O} \rightarrow x \mathrm{I}_2+y \mathrm{MnO}_2+z \overline{\mathrm{O}} \mathrm{H}$
If the above equation is balanced with integer coefficients, the value of $z$ is ________.
Explanation:
| Reduction Half | Oxidation Half |
|---|---|
| $2 \mathrm{MnO}_4^{-} \rightarrow 2 \mathrm{MnO}_2$ | $2 \mathrm{I}^{-} \rightarrow \mathrm{I}_2+2 \mathrm{e}^{-}$ |
| $2 \mathrm{MnO}_4^{-}+4 \mathrm{H}_2 \mathrm{O}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{MnO}_2+8 \mathrm{OH}^{-}$ | $6 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+6 \mathrm{e}^{-}$ |
Adding oxidation half and reduction half, net reaction is
$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+6 \mathrm{I}^{-}+4 \mathrm{H}_2 \mathrm{O} \rightarrow 3 \mathrm{I}_2+2 \mathrm{MnO}_2+8 \mathrm{OH}^{-} \\ \Rightarrow \quad & \mathrm{z}=8 \\ \Rightarrow \quad & \text { Ans } 8 \end{aligned}$
The oxidation number of iron in the compound formed during brown ring test for NO$_3^-$ iron is ________.
Explanation:
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5(\mathrm{NO})\right]^{2+} \text {, }$
Oxidation no. of $\mathrm{Fe}=+1$
See the following chemical reaction:
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{XH}^{+}+6 \mathrm{~F}_{e}^{2+} \rightarrow \mathrm{YCr}^{3+}+6 \mathrm{~F}_{e}^{3+}+\mathrm{Z} \mathrm{H}_{2} \mathrm{O}$
The sum of $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ is ___________
Explanation:
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow \mathrm{YCr}^{3+} + 6 \mathrm{Fe}^{3+} + \mathrm{ZH}_{2} \mathrm{O}$
Let's balance the reaction step by step:
1. Balance the chromium atoms:
There are 2 chromium atoms on the left side of the equation, so we need 2 chromium atoms on the right side. Thus, Y = 2.
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + \mathrm{ZH}_{2} \mathrm{O}$
2. Balance the oxygen atoms:
There are 7 oxygen atoms on the left side of the equation, so we need 7 oxygen atoms on the right side. Since each water molecule (H₂O) contains one oxygen atom, Z = 7.
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$
3. Balance the hydrogen atoms and charges:
There are 14 hydrogen atoms on the right side of the equation, so we need 14 hydrogen atoms on the left side. Thus, X = 14.
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$
Now the chemical reaction is balanced:
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}$
The sum of X, Y, and Z is:
X + Y + Z = 14 + 2 + 7 = 23
So, the sum of X, Y, and Z is 23.
In alkaline medium, the reduction of permanganate anion involves a gain of __________ electrons.
Explanation:
In alkaline medium, the reduction of permanganate anion involves a gain of 3 electrons.
The balanced chemical equation for the reduction of permanganate anion in alkaline medium is:
$ \begin{equation} \ce{MnO4^{-}(aq) + 3 e^{-} + 2 H2O(l) -> MnO2(s) + 4 OH^{-}(aq)} \end{equation}$In this reaction, the permanganate anion (MnO4-) is reduced to manganese dioxide (MnO2) by gaining 3 electrons. The water molecules (H2O) are oxidized by losing 2 electrons. The hydroxide ions (OH-) are spectator ions that do not participate in the redox reaction.
The reduction of permanganate anion in alkaline medium is a useful reaction for a variety of analytical and synthetic purposes. For example, it can be used to quantitatively determine the concentration of an analyte in a solution, or to synthesize a variety of organic compounds.
The sum of oxidation state of the metals in $\mathrm{Fe}(\mathrm{CO})_{5}, \mathrm{VO}^{2+}$ and $\mathrm{WO}_{3}$ is ___________.
Explanation:
To find the sum of the oxidation states of the metals in $\mathrm{Fe(CO)}_{5}$, $\mathrm{VO}^{2+}$, and $\mathrm{WO}_{3}$, we first need to find the oxidation states of the metals in each compound.
$\mathrm{Fe(CO)}_{5}$: The compound $\mathrm{CO}$ is a neutral ligand, so it does not contribute to the oxidation state of the metal. Hence, the oxidation state of Fe in $\mathrm{Fe(CO)}_{5}$ is 0.
$\mathrm{VO}^{2+}$: The compound $\mathrm{O}$ has an oxidation state of -2, and since the overall charge of the compound is +2, the oxidation state of V must be +4 to balance this out.
$\mathrm{WO}_{3}$: The compound $\mathrm{O}$ has an oxidation state of -2, and since there are three oxygen atoms, the total oxidation state contributed by oxygen is -6. To balance this out, the oxidation state of W must be +6.
Therefore, the sum of the oxidation states of the metals in these three compounds is $0 + 4 + 6 = 10$.
In ammonium - phosphomolybdate, the oxidation state of Mo is + ___________
Explanation:
Sum of oxidation states of bromine in bromic acid and perbromic acid is ___________.
Explanation:
The sum of the oxidation states of bromine in bromic acid (HBrO3) and perbromic acid (HBrO4) is +12.
In bromic acid, the oxidation state of bromine is +5, while in perbromic acid, it is +7. The sum of the oxidation states of bromine in both compounds is (+5) + (+7) = +12.
The oxidation state of phosphorus in hypophosphoric acid is + _____________.
Explanation:
Oxidation state is +4
The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is __________ $\times$ 10$^{-2}$ mL (Nearest integer)
Explanation:
The normality of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the solution obtained on mixing $100 \mathrm{~mL}$ of $0.1 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$ with $50 \mathrm{~mL}$ of $0.1 \,\mathrm{M}\, \mathrm{NaOH}$ is _______________ $\times 10^{-1} \mathrm{~N}$. (Nearest Integer)
Explanation:
No. of equivalents of $\mathrm{NaOH}=50 \times 0.1=5$
No. of equivalents of $\mathrm{H}_2 \mathrm{SO}_4$ left $=20-5=15$
$ \begin{aligned} &\Rightarrow 150 \times \mathrm{x}=15 \\\\ &\mathrm{x}=\frac{1}{10}=0.1 \mathrm{~N}=1 \times 10^{-1} \mathrm{~N} \end{aligned} $
$20 \mathrm{~mL}$ of $0.02\, \mathrm{M}$ hypo solution is used for the titration of $10 \mathrm{~mL}$ of copper sulphate solution, in the presence of excess of KI using starch as an indicator. The molarity of $\mathrm{Cu}^{2+}$ is found to be ____________ $\times 10^{-2} \,\mathrm{M}$. [nearest integer]
Given : $2 \,\mathrm{Cu}^{2+}+4 \,\mathrm{I}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2}$
$ \mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-} $
Explanation:
$\mathrm{I}_{2}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}$
Milliequivalents of hypo solution $=0.02 \times 20=0.4$
Milliequivalents of $\mathrm{Cu}^{2+}$ in $10 \mathrm{~mL}$ solution =
Milliequivalents of $\mathrm{I}_{2}=$ Milliequivalents of hypo = 0.4
Millimoles of $\mathrm{Cu}^{2+}$ ions in $10 \mathrm{~mL}=0.4$
Molarity of $\mathrm{Cu}^{2+}$ ions $=\frac{0.4}{10}=0.04 \,\mathrm{M}$
$ =4 \times 10^{-2} \,\mathrm{M} $
0.01 M KMnO4 solution was added to 20.0 mL of 0.05 M Mohr's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4 solution left in the burette after the end point is _____________ mL. (nearest integer)
Explanation:
$\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\mathrm{KMnO}_{4}}=\left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}_{\text{F}}\right)_{\text {Mohr's salt }}$
$0.01 \times 20 \times 5=0.05 \times V \times 1$
Volume required $=20 ~ \mathrm{ml}$
Since initial volume of $\mathrm{KMnO}_{4}$ in burette is $50 ~\mathrm{ml}$.
Hence volume of $\mathrm{KMnO}_{4}$ left in the burette after end point is $30 ~\mathrm{ml}$.
The neutralization occurs when 10 mL of 0.1M acid 'A' is allowed to react with 30 mL of 0.05 M base M(OH)2. The basicity of the acid 'A' is __________.
[M is a metal]
Explanation:
$ \begin{aligned} & \left(\mathrm{M} \times \mathrm{V} \times \mathrm{n}-\text { Factor }_{A}\right.=(\mathrm{M} \times \mathrm{V} \times \mathrm{n}-\text { Factor })_{\mathrm{M}(\mathrm{OH})_{2}} \\\\ & \left[\mathrm{n}-\text { Factor of } \mathrm{M}(\mathrm{OH})_{2}=2\right] \\\\ &0.1 \times 10 \times \mathrm{n} \text {-Factor }= 0.05 \times 30 \times 2 \\\\ & (\mathrm{n}-\text { Factor })_{\mathrm{A}} =3 \end{aligned} $
Hence basicity of acid $A$ is 3.
(Round off to the nearest integer)
Explanation:
or, ${{10 \times 0.05} \over {1000}} \times 5 = {{10 \times M} \over {1000}} \times 2$
$\therefore$ Conc. of oxalic acid solution = 0.125 M
= 0.125 $\times$ 125 g/L = 15.75 g/L
= 1575 $\times$ 10$-$2 g/L
Explanation:
Let Oxidation state of Cr in $CrO_4^{2 - }$ is = x.
$ \therefore $ x + (–2 × 4) = –2
$ \Rightarrow $ x = 6
Explanation:
$NaOH + HCl\buildrel {} \over \longrightarrow NaCl + {H_2}O$
$nf = 1$
$NaC{O_3} + HCl\buildrel {} \over \longrightarrow NaHC{O_3}$
$nf = 1$
Eq of HCl used $ = {n_{NaOH}} \times 1 + {n_{N{a_2}C{O_3}}} \times 1$
$17.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaOH}} + {n_{N{a_2}C{O_3}}}$
2nd end point :
$NaHC{O_3} + HCl\buildrel {} \over \longrightarrow {H_2}C{O_3}$
$1.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaHC{O_3}}} \times 1 = {n_{NaHC{O_3}}}$
0.15 mmol = ${n_{N{a_2}C{O_3}}}$
$0.15 = {n_{N{a_2}C{O_3}}}$
${w_{N{a_2}C{O_3}}} = {{0.15 \times 106 \times {{10}^{ - 3}}} \over {0.4}} \times 100 \times 10$
= 3.975%
$ \simeq $ 4%
(Molar mass of Fe = 56 g mol–1)
Explanation:
nfactor of K2Cr2O7 = 3 $ \times $ 2 = 6
nfactor of FeC2O4 = 1 + 2 = 3
m. eq. of K2Cr2O7 = m. eq. of FeC2O4
$ \Rightarrow $ ${{6 \times 0.02 \times vol} \over {1000}}$ = ${{0.288} \over {144}} \times 3$
$ \Rightarrow $ vol = 50 ml
(The atomic mass of Na is 23 g/mol)
Explanation:
= 23 × 2 + 12 + 48 + 18x
= 46 + 12 + 48 + 18x
= (106 + 18x)
As nfactor in dissolution will be determined from net cationic or anionic charge; which is 2
Eq wt = ${M \over 2}$ = (53 + 9x)
volume = 100 ml = 0.1 Litre
Normality =
$ \Rightarrow $ 0.1= ${{{{1.43} \over {53 + 9x}}} \over {0.1}}$
$ \Rightarrow $ 53 + 9x = 143
$ \Rightarrow $ 9x = 90
$ \Rightarrow $ x = 10
Explanation:
Let oxidation state of Ce
2(+1) + 2x + 7(–2) = 0
$ \Rightarrow $ x = +6
In K2Cr2O7, Transition metal (Cr) present in +6 oxidation state.
KMnO4
(+1) + y + 4(–2) = 0
$ \Rightarrow $ x = +7
In KMnO4, transition metal (Mn) present in +7 oxidation state.
K2FeO4
2(+1) + z + 4(–2) = 0
$ \Rightarrow $x = +6
In K2FeO4, transition metal (Fe) present in +6 oxidation state.
$ \therefore $ x + y + z = 6 + 7 + 6 = 19
One mole of $\mathrm{Cl}_2(\mathrm{~g})$ was passed into 2 L of cold 2 M KOH solution. After the reaction, the concentrations of $\mathrm{Cl}^{-}, \mathrm{ClO}^{-}$and $\mathrm{OH}^{-}$are respectively (assume volume remains constant)
$0.5 \mathrm{M}, 0.5 \mathrm{M}, 0.5 \mathrm{M}$
$1 \mathrm{M}, 1 \mathrm{M}, 1 \mathrm{M}$
$0.5 \mathrm{M}, 0.5 \mathrm{M}, 1 \mathrm{M}$
$0.75 \mathrm{M}, 0.75 \mathrm{M}, 1 \mathrm{M}$
The oxidation state of chromium in the final product formed in the reaction between KI and acidified $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution is :
+2
+3
+4
+6
0.1 M solution of KI reacts with excess of $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{KIO}_3$ solutions. According to equation
$ 5 \mathrm{I}^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O} $
Identify the correct statements :
(A) 200 mL of KI solution reacts with 0.004 mol of $\mathrm{KIO}_3$
(B) 200 mL of KI solution reacts with 0.006 mol of $\mathrm{H}_2 \mathrm{SO}_4$
(C) 0.5 L of KI solution produced 0.005 mol of $\mathrm{I}_2$
(D) Equivalent weight of $\mathrm{KIO}_3$ is equal to ( $\frac{\text { Molecular weight }}{5}$ )
Choose the correct answer from the options given below :
(C) and (D) only
(A) and (B) only
(B) and (C) only
(A) and (D) only
Given below are two statements:
Statement I: In the oxalic acid vs $\mathrm{KMnO}_4$ (in the presence of dil $\mathrm{H}_2 \mathrm{SO}_4$ ) titration the solution needs to be heated initially to $60^{\circ} \mathrm{C}$, but no heating is required in Ferrous ammonium sulphate (FAS) vs $\mathrm{KMnO}_4$ titration (in the presence of dil $\mathrm{H}_2 \mathrm{SO}_4$)
Statement II: In oxalic acid vs $\mathrm{KMnO}_4$ titration, the initial formation of $\mathrm{MnSO}_4$ takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs $\mathrm{KMnO}_4$, heating oxidizes $\mathrm{Fe}^{2+}$ into $\mathrm{Fe}^{3+}$ by oxygen of air and error may be introduced in the experiment.
In the light of the above statements, choose the correct answer from the options given below
Match the List - I with List - II
| List - I (Redox Reaction) |
List - II (Type of Redox Reaction) |
||
|---|---|---|---|
| (A) | $\begin{aligned} & \mathrm{CH}_{4(\mathrm{~g})}+2 \mathrm{O}_{2(\mathrm{~g})} \xrightarrow{\Delta} \mathrm{CO}_{2(\mathrm{~g})} \\ & +2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\end{aligned}$ | (I) | Disproportionation reaction |
| (B) | $2 \mathrm{NaH}_{(\mathrm{s})} \xrightarrow{\Delta} 2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})}$ | (II) | Combination reaction |
| (C) | $\begin{aligned} & \mathrm{V}_2 \mathrm{O}_{5(\mathrm{~s})}+5 \mathrm{Ca}_{(\mathrm{s})} \xrightarrow{\Delta} 2 \mathrm{~V}_{(\mathrm{s})} \\ & +5 \mathrm{CaO}_{(\mathrm{s})}\end{aligned}$ | (III) | Decomposition reaction |
| (D) | $\begin{aligned} & 2 \mathrm{H}_2 \mathrm{O}_{2(\mathrm{aq})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}+ \\ & \mathrm{O}_{2(\mathrm{~g})}\end{aligned}$ | (IV) | Displacement reaction |
Choose the correct answer from the options given below:
Which of the following oxidation reactions are carried out by both $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and $\mathrm{KMnO}_4$ in acidic medium?
A. $\Gamma^{-} \rightarrow \mathrm{I}_2$
B. $\mathrm{S}^{2-} \rightarrow \mathrm{S}$
C. $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}$
D. $\mathrm{I}^{-} \rightarrow \mathrm{IO}_3^{-}$
E. $\mathrm{S}_2 \mathrm{O}_3{ }^{2-} \rightarrow \mathrm{SO}_4^{2-}$
Choose the correct answer from the options given below :
The species which does not undergo disproportionation reaction is :
Thiosulphate reacts differently with iodine and bromine in the reactions given below:
$\begin{aligned} & 2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-} \\ & \mathrm{S}_2 \mathrm{O}_3^{2-}+5 \mathrm{Br}_2+5 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{SO}_4^{2-}+4 \mathrm{Br}^{-}+10 \mathrm{H}^{+} \end{aligned}$
Which of the following statement justifies the above dual behaviour of thiosulphate?
The number of ions from the following that are expected to behave as oxidising agent is :
$\mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Tl}^{3+}, \mathrm{Pb}^{4+}, \mathrm{Tl}^{+}$
Match List I with List II.
| LIST I Reaction |
LIST II Type of redox reaction |
||
|---|---|---|---|
| A. | $ \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})} $ |
I. | Decomposition |
| B. | $ 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{PbO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} $ |
II. | Displacement |
| C. | $ 2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{NaOH}_{(\mathrm{aq} .)}+\mathrm{H}_{2(\mathrm{~g})} $ |
III. | Disproportionation |
| D. | $ 2 \mathrm{NO}_{2(\mathrm{~g})}+2^{-} \mathrm{OH}(\text { aq. }) \rightarrow \mathrm{NO}_{2(\mathrm{aq} .)}^{-}+\mathrm{NO}_{3(\text { aq. })}^{-}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} $ |
IV. | Combination |
Choose the correct answer from the options given below :
When $\mathrm{MnO}_2$ and $\mathrm{H}_2 \mathrm{SO}_4$ is added to a salt $(\mathrm{A})$, the greenish yellow gas liberated as salt (A) is :
(A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$
(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2
\mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
(C) $2 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$
(D) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$
Choose the correct answer from the options given below :
$ \mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O} $
$\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ and $\mathrm{A}$ are respectively are :
Chlorine undergoes disproportionation in alkaline medium as shown below :
$\mathrm{aCl}_{2(\mathrm{~g})}+\mathrm{b} \mathrm{OH}_{(\mathrm{aq})}^{-} \rightarrow \mathrm{c} \mathrm{ClO}_{(\mathrm{aq)}}^{-}+\mathrm{d} \mathrm{Cl}_{(\mathrm{aq})}^{-}+\mathrm{e} \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$
The values of $a, b, c$ and $d$ in a balanced redox reaction are respectively :
In alkaline medium, $\mathrm{MnO}_4^{-}$ oxidises $\mathrm{I}^{-}$ to
Which of the following cannot function as an oxidising agent?
Given below are two statements:
Statement I : In redox titration, the indicators used are sensitive to change in $\mathrm{pH}$ of the solution.
Statement II : In acid-base titration, the indicators used are sensitive to change in oxidation potential.
In the light of the above statements, choose the most appropriate answer from the options given below
$2 \mathrm{IO}_{3}^{-}+x \mathrm{I}^{-}+12 \mathrm{H}^{+} \rightarrow 6 \mathrm{I}_{2}+6 \mathrm{H}_{2} \mathrm{O}$
What is the value of $x$ ?
Which of the following options are correct for the reaction
$2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow 2 \mathrm{Au}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})$
A. Redox reaction
B. Displacement reaction
C. Decomposition reaction
D. Combination reaction
Choose the correct answer from the options given below:
A compound 'X' is a weak acid and it exhibits colour change at pH close to the equivalence point during neutralization of NaOH with $\mathrm{CH}_{3} \mathrm{COOH}$. Compound 'X' exists in ionized form in basic medium. The compound 'X' is
Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$.
Assertion A: Phenolphthalein is a $\mathrm{pH}$ dependent indicator, remains colourless in acidic solution and gives pink colour in basic medium.
Reason R: Phenolphthalein is a weak acid. It doesn't dissociate in basic medium.
In the light of the above statements, choose the most appropriate answer from the options given below.
Which of the given reactions is not an example of disproportionation reaction?