Explanation:
We start with the balanced redox reaction :
$2 \, \mathrm{KMnO}_4 + 3 \, \mathrm{H}_2\mathrm{SO}_4 + 5 \, \mathrm{H}_2\mathrm{S} \rightarrow \mathrm{K}_2\mathrm{SO}_4 + 2 \, \mathrm{MnSO}_4 + 5 \, \mathrm{S} + 8 \, \mathrm{H}_2\mathrm{O}$
We want to find out the number of moles of water ($x$) produced and the number of moles of electrons ($y$) involved in this reaction.
From the balanced equation, we can see that 8 moles of water are produced from the reaction. So, we can say :
$x = 8$
Hydrogen sulfide ($\mathrm{H}_2\mathrm{S}$) gets oxidized to sulfur ($\mathrm{S}$) in this reaction. Each molecule of $\mathrm{H}_2\mathrm{S}$ loses 2 electrons during this process (as sulfur has an oxidation state of -2 in $\mathrm{H}_2\mathrm{S}$ and 0 in $\mathrm{S}$).
So, for every mole of $\mathrm{H}_2\mathrm{S}$, 2 moles of electrons are involved. And since 5 moles of $\mathrm{H}_2\mathrm{S}$ are reacting, the total number of moles of electrons involved is $5 \times 2 = 10$. So, we can say :
$y = 10$
We are asked to find the sum of $x$ and $y$, so we add these two values together to get:
$x + y = 8 + 10 = 18$
Therefore, the value of $(x + y)$ is 18.
[Use, molar mass $\left(\mathrm{g} ~\mathrm{mol}^{-1}\right): \mathrm{H}=1, \mathrm{C}=12, \mathrm{O}=16, \mathrm{Si}=28, \mathrm{Cl}=35.5$ ]
Explanation:
No. of moles $=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{516}{129}=4$
$\begin{aligned} & \therefore \text { Percentage yield }=\frac{75}{100}=0.75 \\\\ & \therefore \text { Mole formed of cyclic tetramer }=0.75 \\\\ & \therefore \text { Weight }=0.75 \times 296=222 \mathrm{~g}\end{aligned}$
The treatment of an aqueous solution of $3.74 \mathrm{~g}$ of $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ with excess KI results in a brown solution along with the formation of a precipitate. Passing $\mathrm{H}_{2} \mathrm{~S}$ through this brown solution gives another precipitate $\mathbf{X}$. The amount of $\mathbf{X}$ (in $g$ ) is ___________.
[Given: Atomic mass of $\mathrm{H}=1, \mathrm{~N}=14, \mathrm{O}=16, \mathrm{~S}=32, \mathrm{~K}=39, \mathrm{Cu}=63, \mathrm{I}=127$ ]
Explanation:
Number of moles of $Cu{(N{O_3})_2} = {{3.74} \over {187}} = 0.02$
$\mathop {2Cu{{(N{O_3})}_2}}\limits_{0.02} + 4KI \to C{u_2}{I_2} \downarrow + \mathop {{I_2}}\limits_{0.01} + 4KN{O_3}$
$\mathop {{I_2}}\limits_{0.01} + KI \to \mathop {K{I_3}}\limits_{0.01\,(Brown\,solution)} $
$\mathop {K{I_3}}\limits_{0.01} + {H_2}S \to KI + \mathop S\limits_{0.01\,(X)} \downarrow + 2HI$
Number of moles of sulphur precipitated (X) = 0.01
Mass of sulphur precipitates (X) = 0.01 $\times$ 32 = 0.32 gm
Explanation:
$2MnO_4^ - + {H_2}O + {I^ - }\buildrel {} \over \longrightarrow 2Mn{O_2} + 2O{H^ - } + IO_3^ - $
But in weakly basic solution :
$\mathop {KMn{O_4}}\limits^{ + 7} + \mathop {KI}\limits^{ - 1} \buildrel {} \over \longrightarrow \mathop {Mn{O_2}}\limits^{ + 4} + \mathop {{I_2}}\limits^0 $
Eq. of KMnO4 = Eq. of I2
$4 \times 3 = n \times 2 \Rightarrow n = 6$
| Exp. No. | Vol. of NaOH (mL) |
|---|---|
| 1 | 12.5 |
| 2 | 10.5 |
| 3 | 9.0 |
| 4 | 9.0 |
| 5 | 9.0 |
Explanation:
${H_2}{C_2}{O_4} + 2NaOH\buildrel {} \over \longrightarrow N{a_2}{C_2}{O_4} + 2{H_2}O$
Equivalent of H2C2O4 reacted = Equivalent of NaOH reacted
$ = {{5 \times 2 \times 0.1} \over {1000}} = {{9 \times {M_{(NaOH)}} \times 1} \over {1000}}$
${M_{(NaOH)}} = {1 \over 9} = 0.11$
(Given data : Molar masses of urea and water are 60 g mol$-$1 and 18 g mol$-$1, respectively)
Explanation:
$(M) = {{Number\,of\,moles\,of\,solute \times 1000} \over {Volume\,of\,solution\,(in\,mL)}}$
Also, volume = ${{Mass} \over {Density}}$
Given, mole fraction of urea $({\chi _{urea}})$ = 0.05
Mass of water = 900 g
Density = 1.2 g/cm3
${\chi _{urea}}$ = ${{{n_{urea}}} \over {{n_{urea}} + 50}}$
[$ \because $ Moles of water = ${{900} \over {18}}$ = 50]
$0.05 = $${{{n_{urea}}} \over {{n_{urea}} + 50}}$
$ \Rightarrow $ $19{n_{urea}} = 50$
${n_{urea}} = 2.6315$ moles
${w_{urea}} = {n_{urea}} \times {(M.wt)_{urea}}$
$ = (2.6315 \times 60)g$
$V = {{2.6315 \times 60 + 900} \over {1.2}}$
[$ \because $ $Density = {{Mass\,of\,solution} \over {Volume\,of\,solution}}$]
= 881.57 mL
Now, molarity
= $Number\,of\,moles\,of\,solute \times {{1000} \over {Volume\,of\,solution\,(mL)}}$
$ = {{2.6315 \times 1000} \over {881.57}}$ = 2.98 M
(Given data : Molar mass of water = 18 g mol$-$1)
Explanation:
${S_8} + 48HN{O_3}\buildrel {} \over \longrightarrow 8{H_2}S{O_4} + 48N{O_2} + 16{H_2}O$
1 mole of rhombic sulphur produces = 16 moles of H2O
$ \therefore $ Mass of water = 16 $ \times $ 18 (molar mass of H2O) = 288 g
$MnC{l_2} + {K_2}{S_2}{O_8} + {H_2}O \to KMn{O_4} + {H_2}S{O_4} + HCl$ (equation not balanced).
Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid ($225$ $mg$) was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnC{l_2}$ (in mg) present in the initial solution is ____________.
(Atomic weights in $g\,\,mo{l^{ - 1}}:Mn = 55,Cl = 35.5$ )
Explanation:
Also,
$2MnO_4^ - + 5{C_2}O_4^{2 - } \to 2M{n^{2 + }} + 10C{O_2}$
Hence, $2M{n^{2 + }} \equiv 5{C_2}O_4^{2 - }$
$1MnC{l_2} \equiv 2.5{H_2}{C_2}{O_4}$
Oxalic acid taken = 225 mg
$ = {{225} \over {90}} = 2.5$ millimoles
Hence, MnCl2 = 1 millimole
= (55 + 71) = 126 mg
Explanation:
The universal gas constant, denoted by R, can be calculated using the Avogadro number (NA) and the Boltzmann constant (kB) by the following relationship:
$ R = N_A \times k_B $Given that:
$ N_A = 6.023 \times 10^{23} \text{ mol}^{-1} $ $ k_B = 1.380 \times 10^{-23} \text{ J K}^{-1} $Let's multiply these values to find R:
$ R = (6.023 \times 10^{23} \text{ mol}^{-1}) \times (1.380 \times 10^{-23} \text{ J K}^{-1}) $ $ R = (6.023 \times 1.380) \times (10^{23} \times 10^{-23}) \text{ J mol}^{-1} \text{K}^{-1} $ $ R = 8.31174 \times 10^{0} \text{ J mol}^{-1} \text{K}^{-1} $To determine the number of significant digits in the calculated value of R, we must consider the number of significant digits in the given values of NA and kB.
The value for NA has four significant digits (6.023), and the value for kB also has four significant digits (1.380). When multiplying or dividing numbers, the number of significant digits in the result is determined by the number with the smallest amount of significant digits used in the calculation.
In this case, since both constants have four significant digits, the value of R calculated from their multiplication will also contain four significant digits:
$ R \approx 8.314 \text{ J mol}^{-1} \text{K}^{-1} $Therefore, the calculated value of the universal gas constant R has four significant digits.
Explanation:
Steps to Calculate Molality
- Find the Mass of Solute (Hâ‚‚X):
- Molarity = moles of solute / volume of solution (in liters)
- 3.2 M solution means 3.2 moles of Hâ‚‚X are present in 1 liter of solution.
- Mass of Hâ‚‚X = moles $ \times $ molar mass = 3.2 moles $ \times $ 80 g/mole = 256 g
- Find the Mass of Solvent:
- Assuming no change in volume, the volume of the solution remains 1 liter.
- Density = mass / volume
- Mass of solvent = density $ \times $ volume = 0.4 g/mL $ \times $ 1000 mL = 400 g
- Convert Mass of Solvent to Kilograms:
- 1 kg = 1000 g
- Mass of solvent = 400 g $ \times $ (1 kg / 1000 g) = 0.4 kg
- Calculate Molality:
- Molality = moles of solute / mass of solvent (in kg)
- Molality = 3.2 moles / 0.4 kg = 8 mol/kg
Answer:
The molality of the 3.2 molar solution is 8 mol/kg.
Explanation:
To calculate the volume of the 29.2% (w/w) HCl stock solution needed to prepare a 200 mL solution of 0.4 M HCl, we need to use several steps involving concentration and density conversions.
First, we calculate the mass of HCl that is contained in the 200 mL of a 0.4 M solution:
$ Mass = Molarity \times Volume \times Molecular\ Weight $
$ Mass = 0.4 \ mol/L \times 0.200 \ L \times 36.5 \ g/mol $
Note that we convert the volume from mL to L to match the units of molarity (mol/L).
Now, we calculate it:
$ Mass = (0.4 \times 0.200 \times 36.5) \ g $
$ Mass = 0.08 \times 36.5 \ g $
$ Mass = 2.92 \ g $
The next step is to determine how much of the stock solution is needed to get 2.92 g of HCl. Since the stock solution is 29.2% (w/w) HCl, this means that in every 100 g of stock solution, there is 29.2 g of HCl. We can set up a proportion to find the mass of the stock solution needed:
$ \frac{29.2\ g \ HCl}{100\ g \ stock\ solution} = \frac{2.92\ g \ HCl}{x\ g \ stock\ solution} $
Now we solve for $ x $:
$ x = \frac{2.92\ g \times 100\ g \ stock\ solution}{29.2\ g \ HCl} $
$ x = \frac{292}{29.2} \ g $
$ x = 10\ g $
So, we need 10 g of the stock solution to get 2.92 g of HCl.
The final step is to calculate the volume of the stock solution that has a mass of 10 g. We use the density to convert mass to volume:
$ Volume = \frac{Mass}{Density} $
The density of the stock solution is given as 1.25 g/mL, so:
$ Volume = \frac{10\ g}{1.25\ g/mL} $
$ Volume = 8\ mL $
Therefore, to prepare a 200 mL solution of 0.4 M HCl, you would need to measure out 8 mL of the 29.2% HCl stock solution.
The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of $[Cr{({H_2}O)_5}Cl]C{l_2}$, as silver chloride is close to ____________.
Explanation:
The reaction taking place is
$2AgN{O_3} + [Cr{({H_2}O)_5}Cl]C{l_2} \to 2AgCl + [Cr{({H_2}O)_5}Cl]{(N{O_3})_2}$
Using molarity equation
${(M \times n \times V)_{AgN{O_3}}} = {(M \times n \times V)_{[Cr{{({H_2}O)}_5}Cl]C{l_2}}}$
$0.1 \times 1 \times V = 0.01 \times 2 \times 30 \Rightarrow V = 6$
Explanation:
To find the average titre value, first add up the three measurements provided and then divide by the number of measurements.
The sum of the measurements is:
$ 25.2 \, \text{mL} + 25.25 \, \text{mL} + 25.0 \, \text{mL} = 75.45 \, \text{mL} $Since there are three measurements, divide this sum by 3 to calculate the average:
$ \text{Average titre value} = \frac{75.45 \, \text{mL}}{3} = 25.15 \, \text{mL} $When reporting the average, we must consider the significant figures of the original measurements. The number of significant figures is determined by the least precise measurement, which in this case is 25.0 mL with three significant figures. Therefore, we should report the average value to three significant figures as well.
The average value of 25.15 mL has four significant figures, so we need to round it to three significant figures. However, this is slightly tricky since 25.15 already appears to be rounded to four significant figures. We should consult the original measurements to decide on the best course of action.
Looking at the individual measurements (25.2, 25.25, and 25.0), we should consider the lowest decimal place which they all have in common, which is the first decimal place. The third measurement has no second decimal place, indicating its level of precision. Thus, the number of significant figures for the average titre value should be in line with this level of precision. Since the average calculated is 25.15, when we adjust to the first decimal place for consistent significant figures, the average is 25.1 mL with three significant figures.
$ \text{Corrected Average titre value} = 25.1 \, \text{mL} $Therefore, the number of significant figures in the average titre value is three: 25.1 mL.
Explanation:
Density $=10.5 \mathrm{~g} \mathrm{~cm}^{-3}$
Surface area $=10^{-12} \mathrm{~m}^2$
Volume of one silver atom $=4 / 3 \pi r^3$
$ \because \text { Density }=\frac{\text { Mass }}{\text { Volume }} \Rightarrow \text { Volume }=\frac{\text { Mass }}{\text { Density }} $
$\begin{aligned} & \text { or } \frac{4}{3} \pi r^3=\frac{108}{6.023 \times 10^{23} \times 10.5} \\\\ & r^3=\frac{108 \times 3}{6.023 \times 10^{23} \times 10.5 \times 4 \times 3.14} \\\\ & r^3=0.40 \times 10^{-23}=4 \times 10^{-24} \\\\ & \text { or } r=1.58 \times 10^{-8} \mathrm{~cm} \end{aligned}$
No. of silver atoms on a surface area of $10^{-12} \mathrm{~m}^2$ can be given by $10^{-12}=\mathrm{p} r^2 \times n$
$\begin{aligned} & n=\frac{10^{-12}}{3.14 \times\left(1.58 \times 10^{-10}\right)^2}=0.127 \times 10^8 \\\\ \Rightarrow & n=1.27 \times 10^7 \text { or } x=7\end{aligned}$
| Compound | Weight % of $\mathrm{P}$ | Weight % of $\mathrm{Q}$ |
|---|---|---|
| 1 | 50 | 50 |
| 2 | 44.4 | 55.6 |
| 3 | 40 | 60 |
1 litre of mixture X + excess AgNO3 $ \to $ Y.
1 litre of mixture X + excess BaCl2 $ \to $ Z
No. of moles of Y and Z are
STATEMENT (S): In the titration of Na2CO3 with HCl using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator.
EXPLANATION (E): Two moles of HCl are required for the complete neutralization of one mole of Na2CO3
Explanation:
Given, 1 L of water = 1 kg = 1000 g (because density = 1000 kg m−3).
Therefore, the number of moles of solute present
= ${{1000} \over {18}}$ = 55.55 mol of H2O
So, the molarity is 55.55 mol/1 L = 55.55 M.
Explanation:
$ =\frac{20 \times M_1}{1000} $
$n$-factor of $\mathrm{KMnO}_4$ when it reacts with $\mathrm{MnSO}_4$ is 3.
$\therefore$ Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{MnSO}_4=\frac{20 \times M_1}{1000} \times 3$
$\therefore$ Equivalent of $\mathrm{MnSO}_4$ reacting with $\mathrm{KMnO}_4=\frac{20 \times M_1}{1000} \times 3$
Since $\mathrm{MnSO}_4$ has $n$-factor 2, the mole of $\mathrm{MnSO}_4$ reacting
$ =\frac{20 \times M_1}{1000} \times \frac{3}{2} $
Total mole of $\mathrm{MnO}_2$ produced $=$ mole of $\mathrm{KMnO}_4+$ mole of $ \mathrm{MnSO}_4 $
Equivalent of $\mathrm{Na}_2 \mathrm{C}_2 \mathrm{O}_4$ reacting with $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$
$\therefore$ Equivalent of $\mathrm{MnO}_2=\frac{10 \times 0.2}{1000} \times 2$
Mole of $\mathrm{MnO}_2$ reacting with sodium oxalate $=\frac{10 \times 0.2 \times 2}{1000 \times 2}$
[as $n$-factor for $\mathrm{MnO}_2$ is 2].
Therefore, $\frac{20 \times M_1}{1000}+\left[\frac{20 \times M_1}{1000} \times \frac{3}{2}\right]=\frac{10 \times 0.2}{1000} ; M_1=0.04$
Equivalent of $\mathrm{KMnO}_4$ reacting with $\mathrm{H}_2 \mathrm{O}_2=\frac{20 \times 0.04}{1000} \times 5$ $ =0.004 $
If molarity of $\mathrm{H}_2 \mathrm{O}_2$ is $M_2$, then $=\frac{20 \times M_2 \times 2}{1000}=0.004$
$\therefore M_2=0.1 \mathrm{M}$
The reactions involved are:
$2 \mathrm{KMnO}_4+5 \mathrm{H}_2\mathrm{O}_2+3 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{K}_2\mathrm{SO}_4+2 \mathrm{MnSO}_4+8 \mathrm{H}_2\mathrm{O}+5 \mathrm{O}_2$,
$2 \mathrm{KMnO}_4+3 \mathrm{MnSO}_4+2 \mathrm{H}_2\mathrm{O} \rightarrow 5 \mathrm{MnO}_2+2 \mathrm{H}_2\mathrm{SO}_4+\mathrm{K}_2\mathrm{SO}_4$,
$\mathrm{MnO}_2+\mathrm{Na}_2\mathrm{C}_2\mathrm{O}_4+2 \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MnSO}_4+2 \mathrm{CO}_2+\mathrm{Na}_2\mathrm{SO}_4+2 \mathrm{H}_2\mathrm{O}$.
Explanation:
To determine how many millilitres of 0.5 M $ \text{H}_2\text{SO}_4 $ are needed to dissolve 0.5 g of copper(II) carbonate $ (\text{CuCO}_3) $, we first need to write the chemical reaction between the copper(II) carbonate and sulfuric acid ( $ \text{H}_2\text{SO}_4 $ ):
$ \text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2 $
From this reaction, it's observable that 1 mole of copper(II) carbonate reacts with 1 mole of sulfuric acid to produce 1 mole of copper(II) sulfate, water, and carbon dioxide.
Let's calculate the number of moles of copper(II) carbonate in 0.5 g:
The molar mass of $ \text{CuCO}_3 $ is calculated as:
$ \text{Molar mass of } \text{CuCO}_3 = (\text{Molar mass of Cu}) + (\text{Molar mass of C}) + 3 \times (\text{Molar mass of O}) $
$ \text{Molar mass of } \text{CuCO}_3 = (63.55 \text{ g/mol}) + (12.01 \text{ g/mol}) + 3 \times (16.00\text{ g/mol}) $
$ \text{Molar mass of } \text{CuCO}_3 = 63.55 + 12.01 + 48.00 $
$ \text{Molar mass of } \text{CuCO}_3 = 123.56 \text{ g/mol} $
Now, we'll find out the number of moles in 0.5 g of $ \text{CuCO}_3 $:
$ \text{Number of moles of } \text{CuCO}_3 = \frac{\text{mass of } \text{CuCO}_3}{\text{Molar mass of } \text{CuCO}_3} $
$ \text{Number of moles of } \text{CuCO}_3 = \frac{0.5 \text{ g}}{123.56 \text{ g/mol}} $
$ \text{Number of moles of } \text{CuCO}_3 \approx 0.004046 \text{ mol} $
Since the reaction is a 1:1 molar ratio, the moles of sulfuric acid needed would also be $ 0.004046 \text{ mol} $.
The concentration (C) of a solution is given by the formula:
$ C = \frac{n}{V} $
where:
- $ C $ is the concentration of the solution in moles per liter (M).
- $ n $ is the number of moles of solute.
- $ V $ is the volume of solution in liters.
Rearranging the formula to solve for volume $ V $:
$ V = \frac{n}{C} $
The given concentration of $ \text{H}_2\text{SO}_4 $ is 0.5 M, which means 0.5 moles of $ \text{H}_2\text{SO}_4 $ per liter. Now, let's calculate the volume required for the $ 0.004046 \text{ mol} $ of $ \text{H}_2\text{SO}_4 $:
$ V = \frac{0.004046 \text{ mol}}{0.5 \text{ M}} = \frac{0.004046 \text{ mol}}{0.5 \text{ mol/L}} $
$ V = 0.008092 \text{ L} $
To convert liters to millilitres:
$ V(\text{mL}) = 0.008092 \text{ L} \times 1000 \frac{\text{mL}}{\text{L}} $
$ V(\text{mL}) = 8.092 \text{ mL} $
Therefore, you will need approximately 8.092 mL of 0.5 M $ \text{H}_2\text{SO}_4 $ to completely dissolve 0.5 g of copper(II) carbonate under the assumption that the reaction goes to completion with no side reactions.
Explanation:
Volume of a cylinder = $\pi$r2h
Radius of virus $ = {{150} \over 2} = 75\mathop A\limits^o = 75 \times {10^{ - 8}}$ cm.
$\therefore$ Volume of one virus
$ = {{22} \over 7} \times {(75 \times {10^{ - 8}})^2} \times 5000 \times {10^{ - 8}}$ cm3
$ = 8.8397 \times {10^{ - 17}}$ cm3
Mass of one virus
$ = {{Volume} \over {Specific\,volume}} = {{8.8393 \times {{10}^{ - 17}}\,c{m^3}} \over {0.75\,c{m^3}\,{g^ - }}}$
$ = 1.178 \times {10^{ - 16}}\,g$
Mol. wt. of virus
$ = 1.178 \times {10^{ - 16}} \times 6.02 \times {10^{23}}\,g$ mol$-$1
$ = 7.09 \times {10^7}\,g$ mol$-$1
Explanation:
To calculate the molarity of the sodium thiosulphate solution, we'll need to apply stoichiometry principles. First, we must determine the number of moles of $ \text{KIO}_3 $ as it will relate to the number of moles of $ \text{I}_2 $ liberated during the reaction. Then we will use the volume of the sodium thiosulphate solution to find its molarity.
The reaction for the iodometry where $ \text{I}_2 $ is liberated from $ \text{KIO}_3 $ and reacted with the excess of $ \text{KI} $ is:
$ \text{KIO}_3 + 5 \text{KI} + 6 \text{HCl} \rightarrow 3 \text{I}_2 + 6 \text{KCl} + 3 \text{H}_2\text{O} $
From the balanced equation, we see that 1 mole of $ \text{KIO}_3 $ produces 3 moles of $ \text{I}_2 $.
Now let's find the number of moles of $ \text{KIO}_3 $:
$ \text{Moles of } \text{KIO}_3 = \frac{\text{mass of } \text{KIO}_3}{\text{molar mass of } \text{KIO}_3} $
$ = \frac{0.10 \text{ g}}{214.0 \text{ g/mol}} $
To get the number of moles, we need to perform the division:
$ = \frac{0.10}{214.0} $
$ = 4.6728972 \times 10^{-4} \text{ moles of } \text{KIO}_3 $
Since 1 mole of $ \text{KIO}_3 $ yields 3 moles of $ \text{I}_2 $, we multiply the moles of $ \text{KIO}_3 $ by 3:
$ \text{Moles of } \text{I}_2 = 3 \times 4.6728972 \times 10^{-4} $
$ = 1.4018692 \times 10^{-3} \text{ moles of } \text{I}_2 $
Next, $ \text{I}_2 $ reacts with sodium thiosulphate ($ \text{Na}_2\text{S}_2\text{O}_3 $) in the following stoichiometric reaction:
$ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 $
From the reaction, 1 mole of $ \text{I}_2 $ reacts with 2 moles of sodium thiosulphate. Therefore, the moles of sodium thiosulphate required to react with the $ \text{I}_2 $ produced can be calculated as follows:
$ \text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = 2 \times \text{Moles of } \text{I}_2 $
$ = 2 \times 1.4018692 \times 10^{-3} $
$ = 2.8037384 \times 10^{-3} \text{ moles of } \text{Na}_2\text{S}_2\text{O}_3 $
Now we have to calculate the molarity of the sodium thiosulphate solution. Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:
$ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} $
We have the number of moles of sodium thiosulphate and the volume is given as 45.0 mL which needs to be converted to liters:
$ \text{Volume in liters} = 45.0 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.045 \text{ L} $
Substitute the number of moles and the volume into the molarity formula:
$ M = \frac{2.8037384 \times 10^{-3} \text{ moles}}{0.045 \text{ L}} $
Performing the division gives us the molarity:
$ M = \frac{2.8037384 \times 10^{-3}}{0.045} $
$ = 0.0623053 \text{ M} $
So, the molarity of the sodium thiosulphate solution is approximately $ 0.0623 $ M.
Explanation:
$ \begin{aligned} & \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{I}^{-} \longrightarrow 3 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_3 \mathrm{O}_4=2\right) \\\\ & \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2\left(n \text {-factor of } \mathrm{Fe}_2 \mathrm{O}_3=2\right) \end{aligned} $
$\begin{aligned} \Rightarrow \text { Meq of } \mathrm{I}_2 \text { formed } & =\mathrm{Meq}\left(\mathrm{Fe}_3 \mathrm{O}_4+\mathrm{Fe}_2 \mathrm{O}_3\right) =\text { Meq of hypo required } \\\\ \Rightarrow 2 x+2 y & =11 \times 0.5 \times 5=27.5 ...........(i)\end{aligned}$
Now, total millimol of $\mathrm{Fe}^{2+}$ formed $=3 x+2 y$. In the reaction
$ \begin{array}{ll} \mathrm{Fe}^{2+}+\mathrm{MnO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Mn}^{2+} \\\\ n \text {-factor of } \mathrm{Fe}^{2+}=1 \\\\ \Rightarrow \text { Meq of } \mathrm{MnO}_4^{-}=\text {Meq of } \mathrm{Fe}^{2+} \\\\ \Rightarrow 3 x+2 y=12.8 \times 0.25 \times 5 \times 2=32 ...........(ii) \end{array} $
Solving Eqs. (i) and (ii), we get
$ \begin{aligned} x=4.5 \text { and } y=9.25 \end{aligned} $
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{4.5}{1000} \times 232=1.044 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_3 \mathrm{O}_4 & =\frac{1.044}{3} \times 100=34.80 \% \\\\ \text { Mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{9.25}{1000} \times 160=1.48 \mathrm{~g} \\\\ \% \text { mass of } \mathrm{Fe}_2 \mathrm{O}_3 & =\frac{1.48}{3} \times 100=49.33 \%\end{aligned}$
Explanation:
Glauber's salt is $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}$.
Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4=142$
Molecular weight of $\mathrm{Na}_2 \mathrm{SO}_4 \cdot 10 \mathrm{H}_2 \mathrm{O}=322$
Weight of Glauber's salt taken $=8.0575 \times 10^{-2} \mathrm{~kg}=80.575 \mathrm{~g}$
Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $322 \mathrm{~g}$ of Glauber's salt
$=142 / 322 \mathrm{~g}$
$\therefore$ Weight of anhydrous $\mathrm{Na}_2 \mathrm{SO}_4$ in $80.575 \mathrm{~g}$ of Glauber's salt
$ =\frac{142}{322} \times 80.575=35.53 \mathrm{~g} $
Hence number of moles of $\mathrm{Na}_2 \mathrm{SO}_4$ per $\mathrm{dm}^3$ of solution
$ =\frac{35.53}{142}=0.25 $
So, the molarity of solution is $0.25 \mathrm{M}$
$ \begin{aligned} \text { Density of solution } & =1077.2 \mathrm{~kg} \mathrm{~m}^{-3} \\\\ & =\frac{1077.2 \times 10^3}{10^6} \mathrm{~g} \mathrm{~cm}^{-3}=1.0772 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned} $
$\begin{aligned} & \text { Weight of solution }=\text { volume } \times \text { density } \\\\ & =1000 \times 1.0772 \mathrm{~g}=1077.2 \mathrm{~g} \\\\ & \text { Weight of water }=(1077.2-35.53)=1041.67 \mathrm{~g} \\\\ & \text { Molality of solution }=\frac{0.25}{1041.67} \times 1000=0.24 \mathrm{~m}\end{aligned}$
$\begin{aligned} & \text { Number of moles of water in solution }=\frac{1041.67}{18}=57.87 \\\\ & \text { Mole fraction of } \mathrm{Na}_2 \mathrm{SO}_4=\frac{\text { Moles of } \mathrm{Na}_2 \mathrm{SO}_4}{\text { Total number of moles }} \\\\ & \qquad=\frac{0.25}{0.25+57.87}=0.0043 \text { or } 4.3 \times 10^{-3}\end{aligned}$
Explanation:
To determine the percentage of iron present as Fe(III) in the sample of wustite (Fe0.93O1.00), we need to assume that the rest of the iron that is not Fe(II) is present as Fe(III). This is because wustite, which nominally has the formula FeO, has a non-stoichiometric composition due to the presence of both Fe(II) and Fe(III) ions that maintain charge neutrality in the lattice.
Given the formula Fe0.93O1.00, for every one oxygen atom, there are 0.93 irons. In terms of charge, the oxygen anion has a charge of -2. The formula can be interpreted as having 0.93 moles of Fe per mole of wustite, and we need to balance the charges considering the presence of both Fe(II) and Fe(III).
Fe(II) has a charge of +2, and Fe(III) has a charge of +3. If we let x be the fraction of Fe(III), then the fraction of Fe(II) would be 0.93 - x because the total amount of iron is 0.93 moles of iron (Fe). Now, we can set up a charge balance equation:
Charge from Fe(II) + Charge from Fe(III) = Charge from O
(0.93 - x)(+2) + (x)(+3) = 1(+2)
2(0.93 - x) + 3x = 2
1.86 - 2x + 3x = 2
Combining like terms gives:
1.86 + x = 2
x = 2 - 1.86
x = 0.14
So, 0.14 moles of Fe are in the form of Fe(III) per mole of wustite.
To find the percentage of Fe(III), you divide the number of moles of Fe(III) by the total moles of Fe and multiply by 100%:
$ \text{Percentage of Fe(III)} = \left( \frac{x}{0.93} \right) \times 100\% $
$ \text{Percentage of Fe(III)} = \left( \frac{0.14}{0.93} \right) \times 100\% $
$ \text{Percentage of Fe(III)} = (0.1505) \times 100\% $
$ \text{Percentage of Fe(III)} = 15.05\% $
Therefore, 15.05% of the iron is present in the form of Fe(III) in the sample of wustite Fe0.93O1.00.
Explanation:
The reaction is $\to$
$C{r_2}{(S{O_4})_3}$ is limiting reagent as ${{2.5} \over 1} < {{11.25} \over 3}$.
$\therefore$ $2.5 - x = 0$
$ \Rightarrow x = 2.5$ milimoles
$\therefore$ Moles of $PbS{O_4}$ formed $ = 3x = 3 \times 2.5 = 7.5 \times {10^{ - 3}}$ moles
After $PbS{O_4}$ precipitate formation in the solution $Pb{(N{O_3})_2}$ and $Cr{(N{O_3})_2}$ remains.
$\therefore$ Milimoles of remaining $Pb{(N{O_3})_2}$ is $ = 11.25 - 3x = 11.25 - 3 \times 2.5 = 3.75$
And milimoles of remaining $Cr{(N{O_3})_2}$ is $ = 2x = 2 \times 2.5 = 5$
$\therefore$ Molar concentration or molarity of $Pb{(N{O_3})_2} = {{3.75 \times {{10}^{ - 3}}} \over {{{45 + 25} \over {1000}}}} = 0.054\,M$
And molarity of $Cr{(N{O_3})_2} = {{5 \times {{10}^{ - 3}}} \over {{{70} \over {1000}}}} = 0.071\,M$
Explanation:
$ 2 \mathrm{NaHCO}_3 \longrightarrow \mathrm{Na}_2 \mathrm{CO}_3+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 $
$ \begin{aligned} \text { Moles of } \mathrm{CO}_2 \text { produced } & =\frac{p V}{R T}=\frac{750}{760} \times \frac{123.9}{1000} \times \frac{1}{0.082 \times 298} \\\\ & =5 \times 10^{-3} \end{aligned} $
$\begin{aligned} & \Rightarrow \text { Moles of } \mathrm{NaHCO}_3 \text { in } 2 \mathrm{~g} \text { sample }=2 \times 5 \times 10^{-3}=0.01 \\\\ & \Rightarrow \text { millimol of } \mathrm{NaHCO}_3 \text { in } 1.5 \mathrm{~g} \text { sample } \\\\ & \qquad=\frac{0.01}{2} \times 1.5 \times 1000=7.5\end{aligned}$
Let the $1.5 \mathrm{~g}$ sample contain $x$ millimol $\mathrm{Na}_2 \mathrm{CO}_3$, then
$ \begin{array}{rlrl} & 2 x+7.5 =\text { millimol of } \mathrm{HCl}=15 \\\\ & \Rightarrow x =3.75 \end{array} $
$\begin{aligned} \Rightarrow \text { Mass of } \mathrm{NaHCO}_3 =\frac{7.5 \times 84}{1000}=0.63 \mathrm{~g} \\\\ \text { Mass of } \mathrm{Na}_2 \mathrm{CO}_3=\frac{3.75 \times 106}{1000}=0.3975 \mathrm{~g}\end{aligned}$
$\begin{aligned} \Rightarrow \% \text { mass of } \mathrm{NaHCO}_3 =\frac{0.63}{1.50} \times 100=42 \% \\\\ \% \text { mass of } \mathrm{Na}_2 \mathrm{CO}_3 =\frac{0.3975}{1.5} \times 100=26.5 \%\end{aligned}$
$\begin{aligned} \% \text { of } \mathrm{Na}_2 \mathrm{SO}_4 \text { in the sample } & =100-(42+26.5) \\\\ & =100-68.5=31.5 \%\end{aligned}$
(Reaction : KIO3 + 2KI + 6HCl $\to$ 3ICl + 3KCl + 3H2O)
Explanation:
First, let's write down the given data for clarity:
- Mass of commercial AgNO3 = 1 g
- Volume of the KI solution used = 50 ml
- Volume of (M/10) KIO3 solution used to titrate excess KI = 50 ml
- Volume of (M/10) KIO3 solution used to titrate 20 ml of KI solution = 30 ml
From the reaction given:
$ KIO_3 + 2KI + 6HCl \to 3ICl + 3KCl + 3H_2O $
It can be seen that 1 mole of KIO3 reacts with 2 moles of KI. The molarity of KIO3 solution is also given as (M/10), which means 0.1 M.
First, let's calculate the moles of KIO3 used to titrate the excess KI present after the precipitation of AgI:
$ \text{Moles of KIO}_3 = \text{Volume} \times \text{Molarity} $
$ = 50 \text{ ml} \times 0.1 \text{ M} $
Since 1 liter is 1000 ml, we convert ml to liters:
$ = \frac{50}{1000} \text{ L} \times \text{0.1 M} $
$ = 0.005 \text{ moles} $
Now, 1 mole of KIO3 reacts with 2 moles of KI; therefore, 0.005 moles of KIO3 will react with:
$ 0.005 \text{ moles KIO}_3 \times 2 \text{ moles KI} / 1 \text{ mole KIO}_3 $
$ = 0.01 \text{ moles KI} $
So, the excess amount of KI after the precipitation of AgI is 0.01 moles.
Next, we need to find out the amount of KI present in the 20 ml of KI solution that reacts with 30 ml of the KIO3 solution.
$ \text{Moles of KIO}_3 (for 20 \text{ ml KI solution}) = \text{Volume} \times \text{Molarity} $
$ = 30 \text{ ml} \times 0.1 \text{ M} $
$ = \frac{30}{1000} \text{ L} \times \text{0.1 M} $
$ = 0.003 \text{ moles KIO}_3 $
This will react with twice the amount of KI:
$ 0.003 \text{ moles KIO}_3 \times 2 \text{ moles KI} / 1 \text{ mole KIO}_3 $
$ = 0.006 \text{ moles KI} $
This amount is present in 20 ml of the KI solution. To find the amount of KI in the initial 50 ml used for the reaction with AgNO3, we set up a proportion, assuming the KI solution is uniform in concentration:
$ \frac{0.006 \text{ moles KI}}{20 \text{ ml}} = \frac{x \text{ moles KI}}{50 \text{ ml}} $
Solving for x:
$ x = \frac{0.006 \text{ moles KI} \times 50 \text{ ml}}{20 \text{ ml}} $
$ x = 0.015 \text{ moles KI} $
From the reaction between AgNO3 and KI:
$ AgNO_3 + KI \to AgI \downarrow + KNO_3 $
You can see that 1 mole of AgNO3 reacts with 1 mole of KI. If 0.01 moles of KI remained after the reaction, the amount of KI that reacted with AgNO3 is:
$ 0.015 \text{ moles KI (total)} - 0.01 \text{ moles KI (excess)} $
$ = 0.005 \text{ moles KI reacted with AgNO}_3 $
Therefore, the moles of AgNO3 in the 1 gram commercial sample is also 0.005 moles, since the molar ratio of AgNO3 to KI is 1:1. Now, let's calculate the mass of 0.005 moles of pure AgNO3:
The molar mass of AgNO3 = Atomic mass of Ag + Atomic mass of N + 3 x Atomic mass of O
= 108 + 14 + 3 x 16 = 170 g/mol
The mass of 0.005 moles of AgNO3 is:
$ 0.005 \text{ moles} \times 170 \text{ g/mol} $
$ = 0.85 \text{ grams} $
Finally, to find the percentage of AgNO3 in the sample, we divide the mass of pure AgNO3 by the mass of the commercial sample and multiply by 100%.
$ \text{Percentage of AgNO}_3 = \frac{0.8493655 \text{ g}}{1 \text{ g}} \times 100\% $
$ = 85\% $
Thus, the sample contains approximately 85 AgNO3.
Find out the molar ratio of Cu2+ to $C_2O_4^{2-}$ in the compound.Write down the balanced redox reactions involved in the above titrations.
Explanation:
93% ${H_2}S{O_4}$ solution weight by volume means in 100 ml solution 93 gm ${H_2}S{O_4}$ present.
Density of solution = 1.84 g/mL
$\therefore$ Weight of solution = 100 $\times$ 1.84 = 184 gm
$\therefore$ Weight of solvent ${H_2}O$ = 184 $-$ 93 = 91 gm
Now,
$\mathrm{Molality ={{Moles\,of\,solute} \over {Weight\,of\,solvent\,in\,Kg}}}$
$ = {{{{93} \over {98}}} \over {{{91} \over {1000}}}} = 10.42$