Some Basic Concepts of Chemistry
$20 \mathrm{~mL}$ 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 $=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{516}{129}=4$
$\begin{aligned} & \therefore \text { Percentage yield }=\frac{75}{100}=0.75 \\\\ & \therefore \text { Mole formed of cyclic tetramer }=0.75 \\\\ & \therefore \text { Weight }=0.75 \times 296=222 \mathrm{~g}\end{aligned}$
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} $
$\% \text { Oxygen }=100-(43.47 + 9.9)$
$=100-53.37 \simeq 46$
$20 \mathrm{~mL}$ of $0.02 \,\mathrm{M} \,\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution is used for the titration of $10 \mathrm{~mL}$ of $\mathrm{Fe}^{2+}$ solution in the acidic medium.
The molarity of $\mathrm{Fe}^{2+}$ solution is __________ $\times \,10^{-2}\, \mathrm{M}$. (Nearest Integer)
Explanation:
$ \begin{aligned} &10 \times 1 \times M=20 \times 6 \times .02 \\\\ &M=24 \times 10^{-2} M \end{aligned} $
$\therefore $ Answer will be 24
A $100 \mathrm{~mL}$ solution of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}$ on treatment with methanol produces $2.24 \mathrm{~mL}$ of a gas at STP. The weight of gas produced is _____________ mg. [nearest integer]
Explanation:
As $2.24 \,\mathrm{ml}$ is formed at STP.
Number of moles of ethane gas produced
$ =\frac{2.24 X}{22.4} $
$=10^{-4}\, \mathrm{ml}$
Mass of ethane produced $=10^{-4} \times 30=3 \times 10^{-}=3\mathrm{mg}$
Chlorophyll extracted from the crushed green leaves was dissolved in water to make $2 \mathrm{~L}$ solution of Mg of concentration $48\, \mathrm{ppm}$. The number of atoms of $\mathrm{Mg}$ in this solution is $x \times 10^{20}$ atoms. The value of $x$ is ___________. (Nearest Integer)
(Given : Atomic mass of $\mathrm{Mg}$ is $24 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$ )
Explanation:
Number of atoms of $\mathrm{Mg}=\frac{96 \times 10^{-3}}{24} \times \mathrm{N}_{\mathrm{A}}$
$ \begin{aligned} &=4 \times 10^{-3} \times 6 \times 10^{23} \\ &=24 \times 10^{20} \end{aligned} $
$x=24$
When 800 mL of 0.5 M nitric acid is heated in a beaker, its volume is reduced to half and 11.5 g of nitric acid is evaporated. The molarity of the remaining nitric acid solution is x $\times$ 10$-$2 M. (Nearest integer)
(Molar mass of nitric acid is 63 g mol$-$1)
Explanation:
Moles of $\mathrm{HNO}_{3}=400 \times 10^{-3} = 0.4 \text { moles }$
Weight of $\mathrm{HNO}_{3}=0.4 \times 63 \mathrm{~g} =25.2 \mathrm{~g} $
Remaining acid $=25.2-11.5 =13.7 \mathrm{~g} $
$ \begin{aligned} M &=\frac{13.7 \times 1000}{400 \times 63} \\ &=\frac{137}{252}=0.54 \\ &=54 \times 10^{-2} \end{aligned} $
56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.
Explanation:
Since $\mathrm{H}_{2}$ is in excess and $20 \mathrm{~L}$ of ammonia gas is produced.
Hence, 2 moles $\mathrm{NH}_{3} \equiv 1$ mole $\mathrm{N}_{2} \quad(v \propto \mathrm{n})$
$ 20 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2} $
Volume of $\mathrm{N}_{2}$ left $=56-10$
$ =46 \mathrm{~L} $
A sample of 4.5 mg of an unknown monohydric alcohol, R-OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is __________ g/mol. [Nearest integer]
Explanation:
moles of alcohol $(\mathrm{ROH}) \equiv$ moles of $\mathrm{CH}_{4}$
At STP, [Assuming STP]
1 mole corresponds to $22.7 \mathrm{~L}$
Hence, $3.1 \mathrm{~mL} \equiv \frac{3.1}{22700} \mathrm{~mol}$
So, moles of alcohol $=\frac{3.1}{22700}$
$\Rightarrow \frac{3.1}{22700}=\frac{4.5 \times 10^{-3}}{\mathrm{M}}$
$M \simeq 33 \mathrm{~g} / \mathrm{mol}$
Blister copper is produced by reaction of copper oxide with copper sulphide.
2Cu2O + Cu2S $\to$ 6Cu + SO2
When 2.86 $\times$ 103 g of Cu2O and 4.77 $\times$ 103 g of Cu2S are used for reaction, the mass of copper produced is _____________ g. (nearest integer)
(Atomic mass of Cu = 63.5 a.m. u, S = 32.0 a.m. u, O = 16.0 a.m. u)
Explanation:
moles of $\mathrm{Cu}_{2} \mathrm{S}=\frac{4.77 \times 10^{3}}{159}=30$
$\mathrm{Cu}_{2} \mathrm{O}$ is limiting reagent
$ \begin{array}{lccc} 2 \mathrm{Cu}_{2} \mathrm{O} & + & \mathrm{Cu}_{2} \mathrm{~S} \longrightarrow & 6 \mathrm{Cu} & + & \mathrm{SO}_{2} \\ 20 && 30 & - & & - \\ - & &20 & 60 & & 10 \end{array} $
Mass of copper $=60 \times 63.5=3810 \mathrm{~g}$
The complete combustion of 0.492 g of an organic compound containing 'C', 'H' and 'O' gives 0.793 g of CO2 and 0.442 g of H2O. The percentage of oxygen composition in the organic compound is ______________. (nearest integer)
Explanation:
Total organic compound = 0.492 gm
Produced CO2 = 0.793 gm
$\therefore$ Moles of CO2 = ${{0.793} \over {44}}$
$\therefore$ Moles of C atoms = ${{0.793} \over {44}}$
$\therefore$ Weight of C atoms = ${{0.793} \over {44}}$ $\times$ 12 = 0.216 g
Produced H2O = 0.442 gm
$\therefore$ Moles of H2O = ${{0.442} \over {18}}$
$\therefore$ Moles of H atoms = ${{0.442} \over {18}}$
$\therefore$ Weight of H atoms = ${{0.442} \over {18}}$ $\times$ 2 = 0.05 g
$\therefore$ Weight of O atoms
= 0.492 $-$ (0.216 + 0.05)
= 0.226 gm
% by mass of oxygen in compound
= ${{0.226} \over {0.492}}$ $\times$ 100 = 46%
116 g of a substance upon dissociation reaction, yields 7.5 g of hydrogen, 60 g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1, 16 and 12, respectively. The data agrees with how many formulae of the following?
A. CH3COOH, B. HCHO, C. CH3OOCH3, D. CH3CHO
Explanation:
Relative atomicities $=\mathrm{H} \Rightarrow 6.5$
$ \begin{aligned} &\mathrm{O} \Rightarrow \frac{51.7}{16}=3.25 \\\\ &\mathrm{C} \Rightarrow \frac{41.8}{12}=3.5 \end{aligned} $
Emperically formula is approx.. $\mathrm{CH}_2 \mathrm{O}$
(A) $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2$ (B) $\mathrm{CH}_2 \mathrm{O}$ relate to this formula.
Two elements A and B which form 0.15 moles of A2B and AB3 type compounds. If both A2B and AB3 weigh equally, then the atomic weight of A is _____________ times of atomic weight of B.
Explanation:
Let atomic weight of A = x
and atomic weight of B = y
$\therefore$ Molar mass of A2B = 2x + y
and molar mass of AB3 = x + 3y
Now, weight of 0.15 moles of A2B = (2x + y) 0.15
and weight of 0.15 moles of AB3 = (x + 3y) 0.15
Given, weight of 0.15 moles of A2B and AB3 are equal.
$\therefore$ (2x + y)0.15 = (x + 3y)0.15
$\Rightarrow$ x = 2y
$ \Rightarrow {x \over y} = 2$
$\therefore$ Atomic weight of A is 2 times of atomic weight of B.
CNG is an important transportation fuel. When 100 g CNG is mixed with 208 g oxygen in vehicles, it leads to the formation of CO2 and H2O and produces large quantity of heat during this combustion, then the amount of carbon dioxide, produced in grams is ____________. [nearest integer]
[Assume CNG to be methane]
Explanation:

Initially,
Mass of CNG(CH4) = 100 gm
$\therefore$ Moles of CH4 = ${{100} \over {16}}$ = 6.25 moles
Mass of O2 = 208 gm
$\therefore$ Moles of O2 = ${{208} \over {32}}$ = 6.5 moles
From reaction you can see,
1 mole of CH4 react with 2 mole of O2
$\therefore$ 6.25 mole of CH4 react with 6.25 $\times$ 2 = 12.5 moles of O2
But here only 6.5 moles of O2 present
So, O2 will act as limiting reagent.
$\therefore$ Produced CO2 will depends on moles of O2.
From 2 moles of O2 1 mole of CO2
$\therefore$ From 6.5 moles of O2 ${{6.5} \over {2}}$ moles of CO2 produced.
$\therefore$ Weight of CO2 = ${{6.5} \over {2}}$ $\times$ 44 = 143 gm
