Some Basic Concepts of Chemistry
Which of the following have same number of significant figures?
A. 0.00253
B. 1.0003
C. 15.0
D. 163
Choose the correct answer from the options given below
The volume of $0.02 ~\mathrm{M}$ aqueous $\mathrm{HBr}$ required to neutralize $10.0 \mathrm{~mL}$ of $0.01 ~\mathrm{M}$ aqueous $\mathrm{Ba}(\mathrm{OH})_{2}$ is (Assume complete neutralization)
$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{AgNO}_{3}$ solution (excess) $\longrightarrow \mathrm{Y}$
$1 \mathrm{~L}$ Solution $(\mathrm{X})+\mathrm{BaCl}_{2}$ solution (excess) $\longrightarrow \mathrm{Z}$
The number of moles of $\mathrm{Y}$ and $\mathrm{Z}$ respectively are
When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A?
What is the mass ratio of ethylene glycol ($\mathrm{C_2H_6O_2}$, molar mass = 62 g/mol) required for making 500 g of 0.25 molal aqueous solution and 250 mL of 0.25 molar aqueous solution?
'25 volume' hydrogen peroxide means
Explanation:
The first step is to write the balanced chemical equation for the reaction between the complex and silver nitrate, as follows:
$\left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) \mathrm{Cl}_2 + 2 \mathrm{AgNO}_3 \longrightarrow \left(\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right) + 2 \mathrm{AgCl} + 2 \mathrm{NO}_3^-$
From the equation, we can see that 1 mole of the complex consumes 2 moles of silver nitrate to form 2 moles of silver chloride. Therefore, the number of millimoles of chloride ions present in the given solution can be calculated as:
$\text{Millimoles of } \mathrm{Cl}^- \text{ ions} = \text{concentration} \times \text{volume} = 0.01\ \mathrm{M} \times 2 \times 10\ \mathrm{mL} = 0.2\ \mathrm{mmol}$
To calculate the volume of 0.1 M silver nitrate required, we can use the formula:
$\text{Millimoles of silver nitrate required} = \text{Millimoles of chloride ions} \times 2 = 0.4\ \mathrm{mmol}$
We can then use the formula:
$\text{Volume of silver nitrate solution} = \frac{\text{Millimoles of silver nitrate required}}{\text{Molarity of silver nitrate solution}}$
Substituting the values, we get:
$\text{Volume of silver nitrate solution} = \frac{0.4\ \mathrm{mmol}}{0.1\ \mathrm{M}} = 4\ \mathrm{mL}$
Therefore, the volume of 0.1 M silver nitrate required for complete precipitation of chloride ions present in 20 mL of 0.01 M $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_{2}$ solution as silver chloride is $\boxed{4\ \mathrm{mL}}$.
$1 \mathrm{~g}$ of a carbonate $\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)$ on treatment with excess $\mathrm{HCl}$ produces $0.01 \mathrm{~mol}$ of $\mathrm{CO}_{2}$. The molar mass of $\mathrm{M}_{2} \mathrm{CO}_{3}$ is __________ $\mathrm{g} ~\mathrm{mol}^{-1}$. (Nearest integer)
Explanation:
$\mathrm{M}_2\mathrm{CO}_3 + 2\mathrm{HCl} \rightarrow 2\mathrm{MCl} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$
From the problem, we know that 1 g of the carbonate produces 0.01 mol of CO2.
To determine the molar mass of the carbonate, we can use the stoichiometry of the reaction:
1 mol of $\mathrm{M}_2\mathrm{CO}_3$ produces 1 mol of CO2.
So, 0.01 mol of CO2 corresponds to 0.01 mol of $\mathrm{M}_2\mathrm{CO}_3$.
Now, we can find the molar mass of $\mathrm{M}_2\mathrm{CO}_3$:
$\frac{\text{mass of }\mathrm{M}_2\mathrm{CO}_3}{\text{moles of }\mathrm{M}_2\mathrm{CO}_3} = \text{molar mass of }\mathrm{M}_2\mathrm{CO}_3$
$\frac{1 \text{ g}}{0.01 \text{ mol}} = 100 \text{ g mol}^{-1}$
So, the molar mass of $\mathrm{M}_2\mathrm{CO}_3$ is approximately 100 g/mol.
An organic compound gives $0.220 \mathrm{~g}$ of $\mathrm{CO}_{2}$ and $0.126 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ on complete combustion. If the $\%$ of carbon is 24 then the $\%$ of hydrogen is __________ $\times 10^{-1}$. ( Nearest integer)
Explanation:
The moles of COâ‚‚ produced can be calculated by dividing the mass of COâ‚‚ produced by its molar mass (44.01 g/mol):
$\text{moles of CO}_{2} = \frac{0.220 \,\mathrm{g}}{44.01 \,\mathrm{g/mol}} = 0.005 \, \mathrm{mol}$
Since each mole of COâ‚‚ contains one mole of carbon, there are 0.005 moles of carbon in the compound.
Now, let's calculate the moles of Hâ‚‚O produced:
$\text{moles of H}_{2}\text{O} = \frac{0.126 \,\mathrm{g}}{18.02 \,\mathrm{g/mol}} = 0.007 \, \mathrm{mol}$
Since each mole of Hâ‚‚O contains two moles of hydrogen, there are 0.014 moles of hydrogen in the compound.
Now, let's find the masses of carbon and hydrogen in the compound:
Mass of carbon = moles of carbon × molar mass of carbon
$\text{Mass of carbon} = 0.005 \, \mathrm{mol} \times 12.01 \,\mathrm{g/mol} = 0.060 \,\mathrm{g}$
Mass of hydrogen = moles of hydrogen × molar mass of hydrogen
$\text{Mass of hydrogen} = 0.014 \, \mathrm{mol} \times 1.008 \,\mathrm{g/mol} = 0.014 \,\mathrm{g}$
Now, we are given that the percentage of carbon is 24%. Let's find the total mass of the compound:
$\text{Total mass of compound} = \frac{\text{Mass of carbon}}{\% \text{ of carbon}} = \frac{0.060 \,\mathrm{g}}{0.24} = 0.250 \,\mathrm{g}$
Finally, let's find the percentage of hydrogen:
$\% \text{ of hydrogen} = \frac{\text{Mass of hydrogen}}{\text{Total mass of compound}} \times 100 = \frac{0.014 \,\mathrm{g}}{0.250 \,\mathrm{g}} \times 100 = 5.6$
So, the percentage of hydrogen is 5.6% or 56 × 10-1.
$20 \mathrm{~mL}$ of calcium hydroxide was consumed when it was reacted with $10 \mathrm{~mL}$ of unknown solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$. Also $20 \mathrm{~mL}$ standard solution of $0.5 ~\mathrm{M} ~\mathrm{HCl}$ containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of $35.5 \mathrm{~mL}$ whereas the burette showed $25.5 \mathrm{~mL}$ initially. The concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is _____________ M. (Nearest integer)
Explanation:
$\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}$
Volume of $\mathrm{Ca}(\mathrm{OH})_2=10 \mathrm{ml}$
Volume of $\mathrm{HCl}=20 \mathrm{ml}$
Concentration of $\mathrm{HCl}=0.5 \mathrm{M}$.
No. of milli moles of $\mathrm{HCl}=10$
No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=5$.
i.e. $\mathrm{M}_{\mathrm{Ca}(\mathrm{OH})_2}=\frac{\text { no. of milli moles }}{\mathrm{V}(\mathrm{ml})}=\frac{5}{10}$ $=0.5 \mathrm{M}$.
Reaction with $\mathrm{H}_2 \mathrm{SO}_4$:
$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O} \text {. }$
No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=20 \times 0.5$ $=10$
i.e. no. of milli moles of $\mathrm{H}_2 \mathrm{SO}_4=10$
$ \begin{aligned} \Rightarrow & \mathrm{M}_{\mathrm{H}_2 \mathrm{SO}_4}=\frac{\text { no, of milli moles }}{\mathrm{V}(\mathrm{ml})} \\\\ & =\frac{10}{10} \\\\ & =1 \mathrm{M} \end{aligned} $
So, the concentration of $\mathrm{H}_2 \mathrm{SO}_4$ is 1 M.
The volume of hydrogen liberated at STP by treating $2.4 \mathrm{~g}$ of magnesium with excess of hydrochloric acid is _________ $\times ~10^{-2} \mathrm{~L}$
Given : Molar volume of gas is $22.4 \mathrm{~L}$ at STP.
Molar mass of magnesium is $24 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
`Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)`
From the equation, we see that 1 mole of magnesium (Mg) produces 1 mole of hydrogen gas (H2).
The molar mass of magnesium (Mg) is given as 24 g/mol. Therefore, 2.4 g of magnesium would correspond to (2.4 g)/(24 g/mol) = 0.1 mol of Mg.
Since 1 mole of Mg produces 1 mole of H2, 0.1 mol of Mg would produce 0.1 mol of H2.
At STP (Standard Temperature and Pressure), the molar volume of a gas is 22.4 L/mol.
Therefore, the volume of 0.1 mol of H2 would be (0.1 mol)*(22.4 L/mol) = 2.24 L.
So, the volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is 2.24 L.
In the terms of the question where the volume is expressed as ______ × 10-2 L, we convert 2.24 L into the desired form, i.e., 2.24 L = 2.24 $ \times $ 102 $ \times $ 10-2 L, so the answer is 224 × 10-2 L.
A solution of sugar is obtained by mixing $200 \mathrm{~g}$ of its $25 \%$ solution and $500 \mathrm{~g}$ of its $40 \%$ solution (both by mass). The mass percentage of the resulting sugar solution is ___________ (Nearest integer)
Explanation:
Let's first calculate the amount of sugar in each of the solutions :
In the 25% sugar solution, the amount of sugar is 25% of 200 g, which equals 50 g.
In the 40% sugar solution, the amount of sugar is 40% of 500 g, which equals 200 g.
Now, let's mix these two solutions together:
The total amount of sugar in the combined solution is 50 g (from the 25% solution) + 200 g (from the 40% solution) = 250 g.
The total weight of the combined solution is 200 g (of the 25% solution) + 500 g (of the 40% solution) = 700 g.
Therefore, the mass percentage of sugar in the combined solution is $ =\frac{250}{700} \times 100 $ = 35.71%
Rounding to the nearest integer gives 36%. So, the mass percentage of the resulting sugar solution is 36%.
$0.5 \mathrm{~g}$ of an organic compound $(\mathrm{X})$ with $60 \%$ carbon will produce __________ $\times 10^{-1} \mathrm{~g}$ of $\mathrm{CO}_{2}$ on complete combustion.
Explanation:
Molar Mass of $\mathrm{CO}_2=12+(16 \times 2)=44$
Mass of Compound $=0.5 \mathrm{~g}$
$ \begin{aligned} &\% \text { of } \mathrm{C} =\frac{\text { Molar mass of } \mathrm{C} \times \text { Mass Of } \mathrm{CO}_2}{\text { Mass Of Compound } \times \text { Molar Mass } \mathrm{Of} ~ \mathrm{CO}_2} \\\\ &\frac{60}{100} =\frac{12 \times x}{0.5 \times 44} \\\\ &1.1 =x \\\\ &x =11 \times 10^{-1} \end{aligned} $
If 5 moles of $\mathrm{BaCl}_{2}$ is mixed with 2 moles of $\mathrm{Na}_{3} \mathrm{PO}_{4}$, the maximum number of moles of $\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}$ formed is ___________ (Nearest integer)
Explanation:
Given the balanced chemical equation:
$3 \mathrm{BaCl}_2 + 2 \mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3\mathrm{(PO}_4)_2 + 6 \mathrm{NaCl}$
We can see that 3 moles of $\mathrm{BaCl}_2$ react with 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$ to produce 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$.
Given that you have 5 moles of $\mathrm{BaCl}_2$ and 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$, let's calculate the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ that can be formed:
From $\mathrm{BaCl}_2$, using the molar ratio from the balanced equation:
$5 \text{ moles } \mathrm{BaCl}_2 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{3 \text{ moles } \mathrm{BaCl}_2} = \frac{5}{3} \text{ moles } \mathrm{Ba}_3\mathrm{(PO}_4)_2$
From $\mathrm{Na}_3\mathrm{PO}_4$, using the molar ratio from the balanced equation:
$2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4} = 1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2$
The limiting reactant is $\mathrm{Na}_3\mathrm{PO}_4$ because it produces 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$, which is less than the $\frac{5}{3}$ moles produced by $\mathrm{BaCl}_2$.
Therefore, the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ formed is 1 mole. Rounding to the nearest integer, the answer is 1.
The molality of a $10 \%(\mathrm{v} / \mathrm{v})$ solution of di-bromine solution in $\mathrm{CCl}_{4}$ (carbon tetrachloride) is '$x$'. $x=$ ____________ $\times 10^{-2} ~\mathrm{M}$. (Nearest integer)
[Given : molar mass of $\mathrm{Br}_{2}=160 \mathrm{~g} \mathrm{~mol}^{-1}$
atomic mass of $\mathrm{C}=12 \mathrm{~g} \mathrm{~mol}^{-1}$
atomic mass of $\mathrm{Cl}=35.5 \mathrm{~g} \mathrm{~mol}^{-1}$
density of dibromine $=3.2 \mathrm{~g} \mathrm{~cm}^{-3}$
density of $\mathrm{CCl}_{4}=1.6 \mathrm{~g} \mathrm{~cm}^{-3}$]
Explanation:
Mass of $90 \mathrm{~mL}^{\mathrm{L}} \mathrm{CCl}_4=90 \times 1.6=144 \mathrm{gm}$
Molality of $\mathrm{Br}_2$ solution in $\mathrm{CCl}_4$
$ \begin{aligned} & =\frac{32 \times 1000}{160 \times 144} \\\\ & =1.39 \mathrm{M} \\\\ & =139 \times 10^{-2} \end{aligned} $
The density of $3 \mathrm{M}$ solution of $\mathrm{NaCl}$ is $1.0 \mathrm{~g} \mathrm{~mL}^{-1}$. Molality of the solution is ____________ $\times 10^{-2} \mathrm{~m}$. (Nearest integer).
Given: Molar mass of $\mathrm{Na}$ and $\mathrm{Cl}$ is $23$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively.
Explanation:
$=363.8 \times 10^{-2}$
Nearest integer $=364$
$2 \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}(\mathrm{g})$
when $12 \mathrm{~g}$ carbon is burnt in $48 \mathrm{~g}$ of oxygen, the volume of carbon monoxide produced is ___________ $\times 10^{-1} \mathrm{~L}$ at STP [nearest integer]
[Given: Assume $\mathrm{CO}$ as ideal gas, Mass of $\mathrm{C}$ is $12 \mathrm{~g} \mathrm{~mol}^{-1}$, Mass of $\mathrm{O}$ is $16 \mathrm{~g} \mathrm{~mol}^{-1}$ and molar volume of an ideal gas at STP is $22.7 \mathrm{~L} \mathrm{~mol}^{-1}$ ]
Explanation:
Limiting reagent is carbon. One mole carbon produces one mole CO. Hence, volume at STP is 227 $ \times $ 10-1 litre.
In the sample of $\mathrm{M}_{0.83} \mathrm{O}_{1.00}$, the percentage of metal ions existing in $+2$ oxidation state is __________ $\%$. (nearest integer)
Explanation:
$ \begin{aligned} & 2 x+3(0.83-x)=2 \\\\ & x=0.49 \\\\ & \% M^{2+}=\frac{0.49}{0.83} \times 100 \\\\ & =59 \% \end{aligned} $
Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of $11.5 \mathrm{~g}$ of zinc with excess $\mathrm{HCl}$ is __________ L (Nearest integer)
(Given : Molar mass of $\mathrm{Zn}$ is $65.4 \mathrm{~g} \mathrm{~mol}^{-1}$ and Molar volume of $\mathrm{H}_{2}$ at $\mathrm{STP}=22.7 \mathrm{~L}$ )
Explanation:
Moles of $\mathrm{Zn}$ used $=\frac{11.5}{65.4}=$ Moles of $\mathrm{H}_2$ evolved
Volume of $\mathrm{H}_2=\frac{11.5}{65.4} \times 22.7 \mathrm{~L}=3.99 \mathrm{~L}$
On complete combustion, $0.492 \mathrm{~g}$ of an organic compound gave $0.792 \mathrm{~g}$ of $\mathrm{CO}_{2}$. The % of carbon in the organic compound is ___________ (Nearest integer)
Explanation:
$ \begin{aligned} & =\frac{12}{44} \times 0.792=0.216 \\\\ & \% \text { of } C \text { in compound }=\frac{0.216}{0.492} \times 100 \\\\ & =43.90 \% \end{aligned} $
The strength of 50 volume solution of hydrogen peroxide is ______ $\mathrm{g} / \mathrm{L}$ (Nearest integer).
Given:Molar mass of $\mathrm{H}_{2} \mathrm{O}_{2}$ is $34 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar volume of gas at $\mathrm{STP}=22.7 \mathrm{~L}$
Explanation:
$\mathrm{H_2O_2\longrightarrow H_2O+\frac{1}{2}O_2}$
$\frac{50}{22.7}$
$\therefore$ Moles of $\mathrm{H_2O_2}$ in solution $=\frac{50}{22.7}\times2$
$\therefore$ Strength $=\frac{\frac{50\times2}{22.7}\times34}{1}=149.78\approx150$
Some amount of dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right)$ is added to $671.141 \mathrm{~mL}$ of chloroform $\left(\mathrm{CHCl}_{3}\right)$ to prepare $2.6 \times 10^{-3} \mathrm{M}$ solution of $\mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{DCM})$. The concentration of $\mathrm{DCM}$ is ___________ ppm (by mass).
Given :
atomic mass : C = 12
H = 1
Cl = 35.5
density of $\mathrm{CHCl}_{3}=1.49 \mathrm{~g} \mathrm{~cm}^{-3}$
Explanation:
Mass of $\mathrm{CHCl_3=671.141\times1.49=1000}$ gm
$2.6 \times {10^{ - 3}} = {{moles\,of\,C{H_2}C{l_2}} \over {0.671141}}$
$\Rightarrow$ moles of $C{H_2}C{l_2} = 1.74496 \times {10^{ - 3}}$
mass of $C{H_2}C{l_2} = 148.32 \times {10^{ - 3}}$ gm
Composition of $C{H_2}C{l_2} = {{148.32 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$
= 148.32 ppm
$\approx$ 148
The volume of HCl, containing 73 g L$^{-1}$, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is __________ mL. (Nearest Integer)
(Given : molar masses of Na, Cl, O, H, are 23, 35.5, 16 and 1 g mol$^{-1}$ respectively.)
Explanation:
$\mathrm{\mathop {Na + {H_2}O}\limits_{0.69\,g} \to \mathop {NaOH + {1 \over 2}{H_2}}\limits_{0.03\,moles}} $
= 0.03 moles
$\therefore 0.03=2\times\mathrm{V}$
$\mathrm{V=\frac{0.03}{2}L}$
= 15 mL
When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO$_2$ was produced. The molar mass of compound is _____________ g mol$^{-1}$ (Nearest integer).
Explanation:
Number of moles of $C{O_2} = {{4.4} \over {44}} = 0.1$
$\therefore$ Number of moles of C in 1 mole of compound = 10
$\therefore$ $120 = {{60} \over {100}} \times (x)$ [where x is molar mass of OC]
Molar mass = 200 g mol$^{-1}$
Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is __________
(Given : Molar mass of A = 84 g mol$^{-1}$)
Explanation:
Mass of carbon in $1 \mathrm{~mol}$ of $(A)=\frac{85.8}{100} \times 84$
$ =72 ~\mathrm{gm} $
Mass of hydrogen in $1 \mathrm{~mol}$ of $(A)=12 ~\mathrm{gm}$
$\therefore$ Number of $\mathrm{H}$-atoms in a molecule of $(\mathrm{A})=12$.
In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is ____________ (Nearest Integer)
(Given : Atomic mass Ba: 137 u, S: 32 u, O: 16 u)
Explanation:
The number of units, which are used to express concentration of solutions from the following is _________
Mass percent, Mole, Mole fraction, Molarity, ppm, Molality
Explanation:
When $\mathrm{Fe_{0.93}O}$ is heated in presence of oxygen, it converts to $\mathrm{Fe_2O_3}$. The number of correct statement/s from the following is ________
A. The equivalent weight of $\mathrm{Fe_{0.93}O}$ is ${{\mathrm{Molecular\,weight}} \over {0.79}}$
B. The number of moles of Fe$^{2+}$ and Fe$^{3+}$ in 1 mole of $\mathrm{Fe_{0.93}O}$ is 0.79 and 0.14 respectively
C. $\mathrm{Fe_{0.93}O}$ is metal deficient with lattice comprising of cubic closed packed arrangement of O$^{2-}$ ions
D. The % composition of Fe$^{2+}$ and Fe$^{3+}$ in $\mathrm{Fe_{0.93}O}$ is 85% and 15% respectively
Explanation:
According to the charge neutrality,
$ \begin{aligned} & 2 x+3(0.93-x)=2 \Rightarrow 2 x-3 x+2.79=2 \Rightarrow x=0.79 \\\\ & \mathrm{Fe}^{2+} \text { ion }=x=0.79 \\\\ & \mathrm{Fe}^{3+} \text { ion }=0.93-x=0.93-0.79=0.14 \\\\ & \% \text { of } \mathrm{Fe}^{2+} \text { ion }=\frac{0.79}{0.93} \times 100=84.95 \% \approx 85 \% \\\\ & \% \text { of } \mathrm{Fe}^{3+} \text { ion }=\frac{0.14}{0.93} \times 100=15.05 \% \end{aligned} $
As it is a metal deficiency defect,
so, 1 mol of $\mathrm{Fe}_{0.93} \mathrm{O}$ contains $1 \mathrm{~mol}$ of oxide ions and $0.93 \mathrm{~mol}$ of iron ions as $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$.
So, moles of $\mathrm{Fe}^{2+}=0.79$
and moles of $\mathrm{Fe}^{3+}=0.14$
$ \text { The equivalent weight }=\frac{\text { Molecular weight }}{n \text {-factor }} $
$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$
For one $\mathrm{Fe}^{2+}, n$-factor $=1$
$\therefore $ For $0.79 \,\mathrm{Fe}^{2+}, n$-factor $=0.79$
Out of $0.93 \mathrm{~mol}$, there are $0.79 \mathrm{~mol} \,\mathrm{Fe}^{2+}$ ions are present.
5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _____________
Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol$^{-1}$ respectively
Explanation:
$\Rightarrow M \times V=500 \times .1$
$\Rightarrow \frac{5}{40} \times \frac{1000}{450} \times V=500 \times 0.1$
$\mathrm{V}=180 \mathrm{~mL}$
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 $=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{516}{129}=4$
$\begin{aligned} & \therefore \text { Percentage yield }=\frac{75}{100}=0.75 \\\\ & \therefore \text { Mole formed of cyclic tetramer }=0.75 \\\\ & \therefore \text { Weight }=0.75 \times 296=222 \mathrm{~g}\end{aligned}$
If $20 \mathrm{~g}$ of $\mathrm{CaCO}_3$ is treated with $100 \mathrm{~mL}$ of $20 \%$ $\mathrm{HCl}$ solution, the amount of $\mathrm{CO}_2$ produced is
Consider the reaction
$4 \mathrm{HNO}_{3}(1)+3 \mathrm{KCl}(\mathrm{s}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{KNO}_{3}(\mathrm{~s})$
The amount of $\mathrm{HNO}_{3}$ required to produce $110.0 \mathrm{~g}$ of $\mathrm{KNO}_{3}$ is
(Given: Atomic masses of $\mathrm{H}, \mathrm{O}, \mathrm{N}$ and $\mathrm{K}$ are $1,16,14$ and 39, respectively.)
$ \begin{aligned} &\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+400 \mathrm{~kJ} \\ &\mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}(\mathrm{g})+100 \mathrm{~kJ} \end{aligned} $
When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into 'CO' and the remaining is converted into '$\mathrm{CO}_{2}$'. The heat generated when $0.6 \mathrm{~kg}$ of coal is burnt is _________.
$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} $
$20 \mathrm{~g} \quad ~~~5 \mathrm{~g}$
Consider the above reaction, the limiting reagent of the reaction and number of moles of $\mathrm{NH}_{3}$ formed respectively are :
$250 \mathrm{~g}$ solution of $\mathrm{D}$-glucose in water contains $10.8 \%$ of carbon by weight. The molality of the solution is nearest to
(Given: Atomic Weights are, $\mathrm{H}, 1 \,\mathrm{u} ; \mathrm{C}, 12 \,\mathrm{u} ; \mathrm{O}, 16 \,\mathrm{u}$)
In Carius method of estimation of halogen, $0.45 \mathrm{~g}$ of an organic compound gave $0.36 \mathrm{~g}$ of $\mathrm{AgBr}$. Find out the percentage of bromine in the compound.
(Molar masses : $\mathrm{AgBr}=188 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$)
Hemoglobin contains $0.34 \%$ of iron by mass. The number of Fe atoms in $3.3 \mathrm{~g}$ of hemoglobin is
(Given: Atomic mass of Fe is $56 \,\mathrm{u}, \mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$.)
$\mathrm{SO}_{2} \mathrm{Cl}_{2}$ on reaction with excess of water results into acidic mixture
$\mathrm{SO}_{2} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{HCl}$
16 moles of $\mathrm{NaOH}$ is required for the complete neutralisation of the resultant acidic mixture. The number of moles of $\mathrm{SO}_{2} \mathrm{Cl}_{2}$ used is :
Using the rules for significant figures, the correct answer for the expression ${{0.02858 \times 0.112} \over {0.5702}}$ will be
Production of iron in blast furnace follows the following equation
Fe3O4(s) + 4CO(g) $\to$ 3Fe(l) + 4CO2(g)
when 4.640 kg of Fe3O4 and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is :
[Given : Molar Atomic mass (g mol$-$1) : Fe = 56, Molar Atomic mass (g mol$-$1) : O = 16, Molar Atomic mass (g mol$-$1) : C = 12]
Compound A contains 8.7% Hydrogen, 74% Carbon and 17.3% Nitrogen. The molecular formula of the compound is,
Given : Atomic masses of C, H and N are 12, 1 and 14 amu respectively.
The molar mass of the compound A is 162 g mol$-$1.
A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is:
(Atomic mass : Cl = 35.5 amu, H = 1 amu)
120 g of an organic compound that contains only carbon and hydrogen gives 330 g of CO2 and 270 g of water on complete combustion. The percentage of carbon and hydrogen, respectively are
If a rocket runs on a fuel (C15H30) and liquid oxygen, the weight of oxygen required and CO2 released for every litre of fuel respectively are :
(Given : density of the fuel is 0.756 g/mL)
A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344 mL of $\mathrm{H}_{2}$ gas at STP. The number of alcoholic hydrogens present in compound 'X' is ________.
Explanation:
STP conditions define the volume of 1 mole of any gas as $22.4$ liters or $22400$ mL. The volume of hydrogen gas evolved here is $1.344$ mL, so we can calculate the number of moles of hydrogen gas ($H_2$) using the formula:
$n = \frac{V}{V_m}$
where:
- $n$ is the number of moles,
- $V$ is the volume of the gas, and
- $V_m$ is the molar volume of the gas.
Substituting in the given values:
$n = \frac{1.344 \text{ mL}}{22400 \text{ mL/mol}} = 5.995 \times 10^{-5} \text{ mol}$
Each molecule of $H_2$ contains $2$ atoms of hydrogen. Therefore, the number of moles of hydrogen atoms is twice the number of moles of hydrogen gas:
$n_H = 2 \times n = 2 \times 5.995 \times 10^{-5} \text{ mol} = 1.199 \times 10^{-4} \text{ mol}$
This number of moles of hydrogen represents the number of moles of alcoholic hydrogen atoms in the $1.84$ mg sample of the compound 'X'.
We can determine the number of moles of compound 'X' in the sample by dividing the mass of the sample by the molar mass of the compound:
$n_X = \frac{m}{M}$
where:
- $n_X$ is the number of moles of 'X',
- $m$ is the mass of 'X', and
- $M$ is the molar mass of 'X'.
Substituting in the given values:
$n_X = \frac{1.84 \text{ mg}}{92 \text{ g/mol}} = \frac{1.84 \times 10^{-3} \text{ g}}{92 \text{ g/mol}} = 2.0 \times 10^{-5} \text{ mol}$
The number of alcoholic hydrogens per molecule of 'X' is then given by the ratio of the number of moles of hydrogen to the number of moles of 'X':
$\frac{n_H}{n_X} = \frac{1.199 \times 10^{-4} \text{ mol}}{2.0 \times 10^{-5} \text{ mol}} = 6$
So, compound 'X' contains $6$ alcoholic hydrogens.
2L of 0.2M H2SO4 is reacted with 2L of 0.1M NaOH solution, the molarity of the resulting product Na2SO4 in the solution is _________ millimolar. (Nearest integer)
Explanation:
Molarity of $\mathrm{Na}_2 \mathrm{SO}_4$ is $\frac{0.1}{4}=0.025 \,\mathrm{M}$
$=25 \,\mathrm{mM}$
In the given reaction,
$X+Y+3 Z \leftrightarrows X YZ_{3}$
if one mole of each of $X$ and $Y$ with $0.05 \mathrm{~mol}$ of $Z$ gives compound $X Y Z_{3}$. (Given : Atomic masses of $X, Y$ and $Z$ are 10, 20 and 30 amu, respectively.) The yield of $X YZ_{3}$ is _____________ g. (Nearest integer)
Explanation:
Limiting reagent is $Z=\frac{0.05}{3}=.016$
3 moles of $Z \rightarrow 1$ mole of $X Y Z_{3}$
$0.05$ mole of $Z \rightarrow \frac{1}{3} \times 0.05$ mole of $X Y Z_{3}$
M.wt. of $\mathrm{XYZ}_{3}=10+20+90$
$=120\, \mathrm{amu}$
Wt. of $X Y Z_{3}=\frac{.05}{3} \times 120$
$=2 \mathrm{~g}$
On complete combustion of $0.492 \mathrm{~g}$ of an organic compound containing $\mathrm{C}, \mathrm{H}$ and $\mathrm{O}$, $0.7938 \mathrm{~g}$ of $\mathrm{CO}_{2}$ and $0.4428 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$ was produced. The % composition of oxygen in the compound is ___________.
Explanation:
$ \begin{aligned} &=\frac{2}{18} \times \frac{0.4428}{0.492} \times 100 \\\\ &=0.11 \times 0.9 \times 100 \\\\ &=0.099 \times 100=9.9 \end{aligned} $
$\%$ of $C=\frac{12}{44} \times \frac{0.7938}{0.492} \times 100$
$ \begin{aligned} &=0.27 \times 1.61 \times 100 \\\\ &=43.47 \end{aligned} $
$\% \text { Oxygen }=100-(43.47 + 9.9)$
$=100-53.37 \simeq 46$