Chapter 1
A Complete Resource Book in Chemistry for JEE Main · 135 exercises
Problem 132
The number of oxalic acid molecules in \(100 \mathrm{ml}\) of \(0.02\) \(\mathrm{N}\) oxalic acid solution is (a) \(6.023 \times 10^{22}\) (b) \(10^{-3}\) (c) \(6.022 \times 10^{20}\) (d) none of these
5 step solution
Problem 133
In the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}\) (I) when \(1 \mathrm{~mol}\) of ammonia and \(1 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) are made to react to completion then (a) \(1.0 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is produced (b) \(1.0 \mathrm{~mol}\) of \(\mathrm{NO}\) will be produced (c) all the ammonia will be consumed (d) all the oxygen will be consumed
3 step solution
Problem 134
Pressure in a mixture of \(4 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(2 \mathrm{~g}\) of \(\mathrm{H}_{2}\) confined in a container of 1 litre capacity at \(0{ }^{\circ} \mathrm{C}\) is (a) \(25.2 \mathrm{~atm}\) (b) \(35.6 \mathrm{~atm}\) (c) \(15.4 \mathrm{~atm}\) (d) \(48.2 \mathrm{~atm}\)
4 step solution
Problem 135
What is the volume (in litres) of \(\mathrm{CO}_{2}\) liberated at STP, when \(2.12\) gram of sodium carbonate (mol. wt \(=106\) ) s treated with excess dilute HCl? (a) \(11.2\) (b) \(2.12\) c) \(0.448\) (d) \(4.26\)
5 step solution
Problem 136
Percentage of Se in peroxidase anhydrous enzyme is \(0.5 \%\) by weight (at. wt = 78.4) then minimum molecular weight of peroxidase anhydrous enzymes is (a) \(1.568 \times 10^{3}\) (b) \(1.568 \times 10^{4}\) (c) \(25.68\) (d) \(4.316 \times 10^{4}\)
5 step solution
Problem 139
The decomposition of a certain mass of \(\mathrm{CaCO}_{3}\) gave \(11.2 \mathrm{dm}^{3}\) of \(\mathrm{CO}_{2}\) gas at STP. The mass of KOH required to completely neutralize the gas is (a) \(56 \mathrm{~g}\) (b) \(28 \mathrm{~g}\) (c) \(42 \mathrm{~g}\) (d) \(20 \mathrm{~g}\)
4 step solution
Problem 140
If 1 mole of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is reacted with 1 mole of \(\mathrm{X}(\mathrm{OH})_{2}\) as: \(\mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{X}(\mathrm{OH})_{2} \longrightarrow \mathrm{XHPO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) then (a) The equivalent weight of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is \(\frac{98}{3}\). (b) The equivalent weight of base is \(\frac{\text { Molecular mass }}{2}\) (c) 1 mole of \(\mathrm{X}(\mathrm{OH})_{2}\) more is required for complete neutralization of \(\mathrm{XHPO}_{4}\).
4 step solution
Problem 141
Which of the following contain the same number of molecules? (a) \(0.1 \mathrm{~mole}\) of \(\mathrm{CO}_{2}\) (b) \(3.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (c) \(0.1 \mathrm{~g}\) atom of Helium gas (d) \(11.2 \mathrm{~L}\) of \(\mathrm{SO}_{2}\) at S.T.P
5 step solution
Problem 146
Which of the following statement is/are correct? (a) \(\mathrm{CrI}_{3}+\mathrm{KOH}+\mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2} \mathrm{CrO}_{4}+\mathrm{KCl}+\mathrm{KIO}_{4}+\mathrm{H}_{2} \mathrm{O}\) So, in balanced chemical reaction coefficient of \(\mathrm{KOH}\) is 64 . (b) If \(\mathrm{Zn}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\) \(\mathrm{MnSO}_{4}+\mathrm{H}_{2} ;\) then equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{KMnO}_{4}+\) equivalents of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O}, \mathrm{n}\)-factor of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is 12 . (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+}+\mathrm{CO}_{2}+\mathrm{NO}_{3}^{-}, \mathrm{n}\)-factor for \([\mathrm{Fe}(\mathrm{CN})]^{3-}\) is 60 .
4 step solution
Problem 148
\(100 \mathrm{ml}\) of \(0.06 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(50 \mathrm{~mL}\) of \(0.06\) \(\mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\). After the reaction is complete (a) \(0.003 \mathrm{M}\) of excess of \(\mathrm{Ca}^{2+}\) will remain in excess. (b) \(0.003\) moles of calcium oxalate will get precipitated (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is excess reagent. (d) \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) is limiting reagent.
5 step solution
Problem 149
Which of the following contains same number of molecules ? (a) \(3.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (b) \(0.1\) mole of \(\mathrm{NO}_{2}\) (c) \(0.1 \mathrm{~g}\) atom of Ar gas (d) \(11.2 \mathrm{~L}\) of \(\mathrm{CO}_{2}\) at S.T.P.
5 step solution
Problem 150
Oleum is considered as a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is obtained by passing \(\mathrm{SO}_{3}\) in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). When \(100 \mathrm{~g}\) sample of oleum is diluted with desired weight of \(\mathrm{H}_{2} \mathrm{O}\), then the total mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) obtained after dilution is known as \% labeling in oleum. For example, an oleum labeled as '109\% \(\mathrm{H}_{2} \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{\prime}\) means the \(109 \mathrm{~g}\) total mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will be formed when \(100 \mathrm{~g}\) of oleum is diluted by \(9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) which combines with all the free \(\mathrm{SO}_{3}\) present in oleum to form \(\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{as} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). What is the \(\%\) of free \(\mathrm{SO}_{3}\) is an oleum that is labeled as \({ }^{4} 104.5 \% \mathrm{H}_{2} \mathrm{SO}_{4}^{\prime} ?\) (a) 30 (b) 10 (c) 20 (d) 40
5 step solution
Problem 151
Oleum is considered as a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is obtained by passing \(\mathrm{SO}_{3}\) in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). When \(100 \mathrm{~g}\) sample of oleum is diluted with desired weight of \(\mathrm{H}_{2} \mathrm{O}\), then the total mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) obtained after dilution is known as \% labeling in oleum. For example, an oleum labeled as '109\% \(\mathrm{H}_{2} \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{\prime}\) means the \(109 \mathrm{~g}\) total mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will be formed when \(100 \mathrm{~g}\) of oleum is diluted by \(9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) which combines with all the free \(\mathrm{SO}_{3}\) present in oleum to form \(\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{as} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). \(9.0 \mathrm{~g}\) water is added into oleum sample labeled as \({ }^{\text {" } 112 \%} \% \mathrm{H}_{2} \mathrm{SO}_{4}\) then the volume of free \(\mathrm{SO}_{3}\) remaining in the solution atl atm pressure and \(0^{\circ} \mathrm{C}\) is (a) \(7.46 \mathrm{~L}\) (b) \(3.73 \mathrm{~L}\) (c) \(11.2 \mathrm{~L}\) (d) \(7.34 \mathrm{~L}\)
6 step solution
Problem 152
Oleum is considered as a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), which is obtained by passing \(\mathrm{SO}_{3}\) in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). When \(100 \mathrm{~g}\) sample of oleum is diluted with desired weight of \(\mathrm{H}_{2} \mathrm{O}\), then the total mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) obtained after dilution is known as \% labeling in oleum. For example, an oleum labeled as '109\% \(\mathrm{H}_{2} \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{\prime}\) means the \(109 \mathrm{~g}\) total mass of pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will be formed when \(100 \mathrm{~g}\) of oleum is diluted by \(9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) which combines with all the free \(\mathrm{SO}_{3}\) present in oleum to form \(\mathrm{H}_{2} \mathrm{SO}_{4} \mathrm{as} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}\). \(1 \mathrm{~g}\) of oleum sample is diluted with water. The solution required \(54 \mathrm{~mL}\) of \(0.4 \mathrm{~N} \mathrm{NaOH}\) for complete neutralization. The \% of free \(\mathrm{SO}_{3}\) in the sample is (a) 20 (b) 52 (c) 26 (d) none of these
6 step solution
Problem 156
The value of \(\mathrm{n}\) (the number of electrons) in the half equation in which \(\mathrm{Cr}(\mathrm{OH})_{4}^{-}\)is oxidized to chromate ion in alkali medium is
5 step solution
Problem 157
On heating \(1.763 \mathrm{~g}\) of hydrated \(\mathrm{BaCl}_{2}\) to dryness, \(1.505 \mathrm{~g}\) of anhydrous salt remained. Number of moles of \(\mathrm{H}_{2} \mathrm{O}\) present in one mole of the hydrated \(\mathrm{BaCl}_{2}\) is [Mol. wt. of anhydrous \(\mathrm{BaCl}_{2}\) is 208\(]\).
5 step solution
Problem 158
Equivalent weight of a metal is \(4.5\) and the molecular weight of the chloride is 80 . The valency of the metal is
6 step solution
Problem 159
\(14.96 \mathrm{~g}\) of a mixture containing \(\mathrm{n}\)-hexane and ethyl alcohol are reacted with sodium metal to give \(200 \mathrm{ml}\) of \(\mathrm{H}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(760 \mathrm{~mm}\) pressure. The percentage of ethyl alcohol in the mixture is \([\mathrm{R}=0.082 \mathrm{~L}\) atm \(\left.\mathrm{mol}^{-1} \mathrm{~K}^{-1}\right]\)
5 step solution
Problem 160
Consider the reaction, \(3 \mathrm{Br}_{2}+6 \mathrm{NaOH} \longrightarrow 5 \mathrm{NaBr}+\) \(\mathrm{NaBrO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\). If \(\mathrm{E}\) and A represent equivalent weight and atomic weight of bromine, then \(5 \mathrm{E}=\mathrm{x} \mathrm{A}\). The value of \(x\) is
6 step solution
Problem 161
Find the number of moles of KCl in \(1000 \mathrm{~mL}\) of \(3 \mathrm{M}\) solution
4 step solution
Problem 162
\(1 \mathrm{~g}\) of an acid \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) required \(0.768 \mathrm{~g}\) of \(\mathrm{KOH}\) for complete neutralization. Determine the basicity of acid.
5 step solution
Problem 163
When the equation \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{I}_{2}\) Is balanced, give the correct coefficient for \(\mathrm{I}^{-}\)(aq)
9 step solution
Problem 164
How many of the following depends upon temperature? Concentration, Normality, molarity, formality molality, mol fraction, \% by mass, \%by strength
5 step solution
Problem 165
Number of atoms in \(560 \mathrm{~g}\) of \(\mathrm{Fe}\) (atomic mass \(56 \mathrm{~g}\) \(\mathrm{mol}^{-1}\) ) is (a) is twice that of \(70 \mathrm{~g} \mathrm{~N}\) (b) is half that of \(20 \mathrm{~g} \mathrm{H}\) (c) both are correct (d) none is correct
9 step solution
Problem 166
What volume of hydrogen gas, at \(273 \mathrm{~K}\) and 1 atm pressure will be consumed in obtaining \(21.6 \mathrm{~g}\) of elemental boron (atomic mass \(=10.8\) ) from the reduction of boron trichloride by hydrogen? \([2003]\) (a) \(89.6 \mathrm{~L}\) (b) \(67.2 \mathrm{~L}\) (c) \(44.8 \mathrm{~L}\) (d) \(22.4 \mathrm{~L}\)
5 step solution
Problem 168
To neutralize completely \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous solution of phosphorus acid, the volume of \(0.1 \mathrm{M}\) aqueous KOH solution required is [2004] (a) \(10 \mathrm{~mL}\) (b) \(40 \mathrm{~mL}\) (c) \(60 \mathrm{~mL}\) (d) \(80 \mathrm{~mL}\)
4 step solution
Problem 169
\(6.02 \times 10^{20}\) molecules of urea are present in \(100 \mathrm{~mL}\) of its solution. The concentration of urea solution is [2004] (a) \(0.02 \mathrm{M}\) (b) \(0.001 \mathrm{M}\) (c) \(0.01 \mathrm{M}\) (d) \(0.1 \mathrm{M}\)
5 step solution
Problem 170
If we consider that \(\frac{1}{6}\), in place of \(\frac{1}{12}\), mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will (a) decrease twice (b) increase two fold (c) remain unchanged (d) be a function of the molecular mass of the substance
4 step solution
Problem 171
How many moles of magnesium phosphate, \(\mathrm{Mg}_{3}\) \(\left(\mathrm{PO}_{4}\right)_{2}\) will contain \(0.25\) mole of oxygen atoms? [2006] (a) \(0.02\) (b) \(3.125 \times 10^{-2}\) (c) \(1.25 \times 10^{-2}\) (d) \(2.5 \times 10^{-2}\)
4 step solution
Problem 172
Density of a \(2.05 \mathrm{M}\) solution of acetic acid in water is \(1.02 \mathrm{~g} / \mathrm{mL}\). The molality of the solution is(a) \(1.14 \mathrm{~mol} \mathrm{~kg}^{-1}\) (b) \(3.28 \mathrm{~mol} \mathrm{~kg}^{-1}\) (c) \(2.28 \mathrm{~mol} \mathrm{~kg}^{-1}\) (d) \(0.44 \mathrm{~mol} \mathrm{~kg}^{-1}\)
7 step solution
Problem 173
The reaction, \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{Al}^{3+}(\mathrm{aq})\) $$ \begin{gathered} +6 \mathrm{Cl}^{-}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g}) \\ \\\ \text { [2007] } \end{gathered} $$ (a) \(33.6 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) is produced regardless of temperature and pressure for every mole of \(\mathrm{Al}\) that reacts (b) \(67.2 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at STP is produced for every mole of Al that reacts (c) \(11.2 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at STP is produced for every mole of HCl (aq) consumed (d) \(6 \mathrm{~L} \mathrm{HCl}\) (aq) is consumed for every \(3 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) produced
5 step solution
Problem 174
The density (in \(\mathrm{g} \mathrm{mL}^{-1}\) ) of a \(3.60 \mathrm{M}\) sulphuric acid solution, that is, \(29 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) (molar mass \(=98 \mathrm{~g}\). \(\mathrm{mol}^{-1}\) ) by mass will be [2007] (a) \(1.88\) (b) \(1.22\) (c) \(1.45\) (d) \(1.64\)
7 step solution
Problem 175
A \(5.2\) molal aqueous solution of methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH}\), is supplied. What is the mole fraction of methyl alcohol in the solution? (a) \(0.86\) (b) \(0.086\) (c) \(0.043\) (d) \(1.0\)
6 step solution
Problem 176
The density of a solution prepared by dissolving \(120 \mathrm{~g}\) of urea (mol. Mass \(=60 \mathrm{u}\) ) in \(1000 \mathrm{~g}\) of water is \(1.15 \mathrm{~g} / \mathrm{mL}\). The molarity of this solutions is: [2012] (a) \(1.02 \mathrm{M}\) (b) \(0.50 \mathrm{M}\) (c) \(2.05 \mathrm{M}\) (d) \(1.78 \mathrm{M}\)
6 step solution
Problem 177
The molarity of a solution obtained by mixing 750 \(\mathrm{mL}\) of \(0.5(\mathrm{M}) \mathrm{HCl}\) with \(250 \mathrm{~mL}\) of \(2(\mathrm{M}) \mathrm{HCl}\) will be [2013] (a) \(1.75 \mathrm{M}\) (b) \(0.975 \mathrm{M}\) (c) \(0.875 \mathrm{M}\) (d) \(1.78 \mathrm{M}\)
5 step solution