228740
How much litres of $\mathrm{CO}_2$ at STP will be formed when $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ reacts with excess of $\mathrm{Na}_2 \mathrm{CO}_3$ ?
1 22.4
2 2.24
3 0.224
4 5.6
Explanation:
$\underset{\substack{1 \text { mol } \\ 100 \mathrm{~mL}}}{\mathrm{H}_2 \mathrm{SO}_4}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \underset{22.4 \mathrm{~L}}{\mathrm{CO}_2}+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}$ 1 mole of sulfuric acid will produce 1 mole (22.42 at STP) of carbon dioxide. $100 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=10$ million mole of $\mathrm{H}_2 \mathrm{SO}_4$ $=10 \times 10^{-3}$ $=10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ $\because 1$ mole of $\mathrm{H}_2 \mathrm{SO}_4$ liberate $\mathrm{CO}_2=22.4 \mathrm{~L}$ $\therefore 10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ liberates $\mathrm{CO}_2$ $\begin{aligned} & =22.4 \times 10^{-2} \\ & =0.224 \mathrm{~L} \end{aligned}$ Thus, 0.01 mole of sulfuric acid will produce 0.01 mole $\left(0.224 \mathrm{~L}\right.$ at STP) of $\mathrm{CO}_2$.
AP-EAMCET-1998
Some Basic Concepts of Chemistry
228741
$\mathrm{KMnO}_4$ oxidizes oxalic acid in acid medium. The number of $\mathrm{CO}_2$ molecules produced as per the balanced equation is:
1 10
2 8
3 6
4 3
Explanation:
When potassium permanganate $\left(\mathrm{KMnO}_4\right)$ oxidize the oxalic acid then following reaction take place during the reaction- $\begin{aligned} 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow & \mathrm{K}_2 \mathrm{SO}_4+ \\ & 2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O} \end{aligned}$ In a balanced chemical equation of oxidation of oxalic acid by $\mathrm{KMnO}_4$ in acidic medium evolve 10 molecules of $\mathrm{CO}_2$.
AP EAMCET (Medical) 1998
Some Basic Concepts of Chemistry
228742
The volume in litres of $\mathrm{CO}_2$ liberated at STP when $10 \mathrm{~g}$ of $90 \%$ pure lime stone is heated completely is:
1 2.016
2 20.16
3 2.24
4 22.4
Explanation:
$\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\underset{\substack{22.4 \mathrm{LatatST} \\ \text { (1mole) }}}{\mathrm{CO}_2}$ $\therefore$ Molecular weight of $\mathrm{CaCO}_3=100 \mathrm{~g} / \mathrm{mole}$ Weight of $10 \mathrm{gm}$ of $90 \%$ pure limestone $=\frac{10 \times 90}{100}$ $=9$ gm pure $\mathrm{CaCO}_3$ $\therefore 9 \mathrm{gm}$ pure $\mathrm{CaCO}_3$ $\mathrm{CO}_2=\frac{22.4 \times 9}{100}=2.016 \mathrm{~L} \text {. }$
AP EAMCET (Medical) -1998 AP-EAMCET-1996
Some Basic Concepts of Chemistry
228743
The total number of valence electrons in $4.2 \mathrm{~g}$ of $\mathrm{N}_3^{-}$ion is ( $\mathrm{N}_{\mathrm{A}}$ is the Avogadro's number)
1 $2.1 \mathrm{~N}_{\mathrm{A}}$
2 $4.2 \mathrm{~N}_{\mathrm{A}}$
3 $1.6 \mathrm{~N}_{\mathrm{A}}$
4 $3.2 \mathrm{~N}_{\mathrm{A}}$
Explanation:
Molecular wt of $\mathrm{N}_3^{-}=3 \times 14=42 \mathrm{~g}$ Moles of $\mathrm{N}_3^{-}$ion $=\frac{4.2}{42}=0.1$ Each nitrogen atom has 5 valence electrons, total number of electrons in $\mathrm{N}_3^{-} \text {ion }=16$ Total number of electrons in 0.1 mole, $4.2 \mathrm{~g} \text { of } \mathrm{N}_3^{-} \text {ion }=0.1 \times 16 \times \mathrm{N}_{\mathrm{A}}=1.6 \mathrm{~N}_{\mathrm{A}}$
228740
How much litres of $\mathrm{CO}_2$ at STP will be formed when $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ reacts with excess of $\mathrm{Na}_2 \mathrm{CO}_3$ ?
1 22.4
2 2.24
3 0.224
4 5.6
Explanation:
$\underset{\substack{1 \text { mol } \\ 100 \mathrm{~mL}}}{\mathrm{H}_2 \mathrm{SO}_4}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \underset{22.4 \mathrm{~L}}{\mathrm{CO}_2}+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}$ 1 mole of sulfuric acid will produce 1 mole (22.42 at STP) of carbon dioxide. $100 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=10$ million mole of $\mathrm{H}_2 \mathrm{SO}_4$ $=10 \times 10^{-3}$ $=10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ $\because 1$ mole of $\mathrm{H}_2 \mathrm{SO}_4$ liberate $\mathrm{CO}_2=22.4 \mathrm{~L}$ $\therefore 10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ liberates $\mathrm{CO}_2$ $\begin{aligned} & =22.4 \times 10^{-2} \\ & =0.224 \mathrm{~L} \end{aligned}$ Thus, 0.01 mole of sulfuric acid will produce 0.01 mole $\left(0.224 \mathrm{~L}\right.$ at STP) of $\mathrm{CO}_2$.
AP-EAMCET-1998
Some Basic Concepts of Chemistry
228741
$\mathrm{KMnO}_4$ oxidizes oxalic acid in acid medium. The number of $\mathrm{CO}_2$ molecules produced as per the balanced equation is:
1 10
2 8
3 6
4 3
Explanation:
When potassium permanganate $\left(\mathrm{KMnO}_4\right)$ oxidize the oxalic acid then following reaction take place during the reaction- $\begin{aligned} 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow & \mathrm{K}_2 \mathrm{SO}_4+ \\ & 2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O} \end{aligned}$ In a balanced chemical equation of oxidation of oxalic acid by $\mathrm{KMnO}_4$ in acidic medium evolve 10 molecules of $\mathrm{CO}_2$.
AP EAMCET (Medical) 1998
Some Basic Concepts of Chemistry
228742
The volume in litres of $\mathrm{CO}_2$ liberated at STP when $10 \mathrm{~g}$ of $90 \%$ pure lime stone is heated completely is:
1 2.016
2 20.16
3 2.24
4 22.4
Explanation:
$\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\underset{\substack{22.4 \mathrm{LatatST} \\ \text { (1mole) }}}{\mathrm{CO}_2}$ $\therefore$ Molecular weight of $\mathrm{CaCO}_3=100 \mathrm{~g} / \mathrm{mole}$ Weight of $10 \mathrm{gm}$ of $90 \%$ pure limestone $=\frac{10 \times 90}{100}$ $=9$ gm pure $\mathrm{CaCO}_3$ $\therefore 9 \mathrm{gm}$ pure $\mathrm{CaCO}_3$ $\mathrm{CO}_2=\frac{22.4 \times 9}{100}=2.016 \mathrm{~L} \text {. }$
AP EAMCET (Medical) -1998 AP-EAMCET-1996
Some Basic Concepts of Chemistry
228743
The total number of valence electrons in $4.2 \mathrm{~g}$ of $\mathrm{N}_3^{-}$ion is ( $\mathrm{N}_{\mathrm{A}}$ is the Avogadro's number)
1 $2.1 \mathrm{~N}_{\mathrm{A}}$
2 $4.2 \mathrm{~N}_{\mathrm{A}}$
3 $1.6 \mathrm{~N}_{\mathrm{A}}$
4 $3.2 \mathrm{~N}_{\mathrm{A}}$
Explanation:
Molecular wt of $\mathrm{N}_3^{-}=3 \times 14=42 \mathrm{~g}$ Moles of $\mathrm{N}_3^{-}$ion $=\frac{4.2}{42}=0.1$ Each nitrogen atom has 5 valence electrons, total number of electrons in $\mathrm{N}_3^{-} \text {ion }=16$ Total number of electrons in 0.1 mole, $4.2 \mathrm{~g} \text { of } \mathrm{N}_3^{-} \text {ion }=0.1 \times 16 \times \mathrm{N}_{\mathrm{A}}=1.6 \mathrm{~N}_{\mathrm{A}}$
228740
How much litres of $\mathrm{CO}_2$ at STP will be formed when $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ reacts with excess of $\mathrm{Na}_2 \mathrm{CO}_3$ ?
1 22.4
2 2.24
3 0.224
4 5.6
Explanation:
$\underset{\substack{1 \text { mol } \\ 100 \mathrm{~mL}}}{\mathrm{H}_2 \mathrm{SO}_4}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \underset{22.4 \mathrm{~L}}{\mathrm{CO}_2}+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}$ 1 mole of sulfuric acid will produce 1 mole (22.42 at STP) of carbon dioxide. $100 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=10$ million mole of $\mathrm{H}_2 \mathrm{SO}_4$ $=10 \times 10^{-3}$ $=10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ $\because 1$ mole of $\mathrm{H}_2 \mathrm{SO}_4$ liberate $\mathrm{CO}_2=22.4 \mathrm{~L}$ $\therefore 10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ liberates $\mathrm{CO}_2$ $\begin{aligned} & =22.4 \times 10^{-2} \\ & =0.224 \mathrm{~L} \end{aligned}$ Thus, 0.01 mole of sulfuric acid will produce 0.01 mole $\left(0.224 \mathrm{~L}\right.$ at STP) of $\mathrm{CO}_2$.
AP-EAMCET-1998
Some Basic Concepts of Chemistry
228741
$\mathrm{KMnO}_4$ oxidizes oxalic acid in acid medium. The number of $\mathrm{CO}_2$ molecules produced as per the balanced equation is:
1 10
2 8
3 6
4 3
Explanation:
When potassium permanganate $\left(\mathrm{KMnO}_4\right)$ oxidize the oxalic acid then following reaction take place during the reaction- $\begin{aligned} 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow & \mathrm{K}_2 \mathrm{SO}_4+ \\ & 2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O} \end{aligned}$ In a balanced chemical equation of oxidation of oxalic acid by $\mathrm{KMnO}_4$ in acidic medium evolve 10 molecules of $\mathrm{CO}_2$.
AP EAMCET (Medical) 1998
Some Basic Concepts of Chemistry
228742
The volume in litres of $\mathrm{CO}_2$ liberated at STP when $10 \mathrm{~g}$ of $90 \%$ pure lime stone is heated completely is:
1 2.016
2 20.16
3 2.24
4 22.4
Explanation:
$\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\underset{\substack{22.4 \mathrm{LatatST} \\ \text { (1mole) }}}{\mathrm{CO}_2}$ $\therefore$ Molecular weight of $\mathrm{CaCO}_3=100 \mathrm{~g} / \mathrm{mole}$ Weight of $10 \mathrm{gm}$ of $90 \%$ pure limestone $=\frac{10 \times 90}{100}$ $=9$ gm pure $\mathrm{CaCO}_3$ $\therefore 9 \mathrm{gm}$ pure $\mathrm{CaCO}_3$ $\mathrm{CO}_2=\frac{22.4 \times 9}{100}=2.016 \mathrm{~L} \text {. }$
AP EAMCET (Medical) -1998 AP-EAMCET-1996
Some Basic Concepts of Chemistry
228743
The total number of valence electrons in $4.2 \mathrm{~g}$ of $\mathrm{N}_3^{-}$ion is ( $\mathrm{N}_{\mathrm{A}}$ is the Avogadro's number)
1 $2.1 \mathrm{~N}_{\mathrm{A}}$
2 $4.2 \mathrm{~N}_{\mathrm{A}}$
3 $1.6 \mathrm{~N}_{\mathrm{A}}$
4 $3.2 \mathrm{~N}_{\mathrm{A}}$
Explanation:
Molecular wt of $\mathrm{N}_3^{-}=3 \times 14=42 \mathrm{~g}$ Moles of $\mathrm{N}_3^{-}$ion $=\frac{4.2}{42}=0.1$ Each nitrogen atom has 5 valence electrons, total number of electrons in $\mathrm{N}_3^{-} \text {ion }=16$ Total number of electrons in 0.1 mole, $4.2 \mathrm{~g} \text { of } \mathrm{N}_3^{-} \text {ion }=0.1 \times 16 \times \mathrm{N}_{\mathrm{A}}=1.6 \mathrm{~N}_{\mathrm{A}}$
228740
How much litres of $\mathrm{CO}_2$ at STP will be formed when $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ reacts with excess of $\mathrm{Na}_2 \mathrm{CO}_3$ ?
1 22.4
2 2.24
3 0.224
4 5.6
Explanation:
$\underset{\substack{1 \text { mol } \\ 100 \mathrm{~mL}}}{\mathrm{H}_2 \mathrm{SO}_4}+\mathrm{Na}_2 \mathrm{CO}_3 \rightarrow \underset{22.4 \mathrm{~L}}{\mathrm{CO}_2}+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}$ 1 mole of sulfuric acid will produce 1 mole (22.42 at STP) of carbon dioxide. $100 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4=10$ million mole of $\mathrm{H}_2 \mathrm{SO}_4$ $=10 \times 10^{-3}$ $=10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ $\because 1$ mole of $\mathrm{H}_2 \mathrm{SO}_4$ liberate $\mathrm{CO}_2=22.4 \mathrm{~L}$ $\therefore 10^{-2}$ moles of $\mathrm{H}_2 \mathrm{SO}_4$ liberates $\mathrm{CO}_2$ $\begin{aligned} & =22.4 \times 10^{-2} \\ & =0.224 \mathrm{~L} \end{aligned}$ Thus, 0.01 mole of sulfuric acid will produce 0.01 mole $\left(0.224 \mathrm{~L}\right.$ at STP) of $\mathrm{CO}_2$.
AP-EAMCET-1998
Some Basic Concepts of Chemistry
228741
$\mathrm{KMnO}_4$ oxidizes oxalic acid in acid medium. The number of $\mathrm{CO}_2$ molecules produced as per the balanced equation is:
1 10
2 8
3 6
4 3
Explanation:
When potassium permanganate $\left(\mathrm{KMnO}_4\right)$ oxidize the oxalic acid then following reaction take place during the reaction- $\begin{aligned} 2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \longrightarrow & \mathrm{K}_2 \mathrm{SO}_4+ \\ & 2 \mathrm{MnSO}_4+10 \mathrm{CO}_2+8 \mathrm{H}_2 \mathrm{O} \end{aligned}$ In a balanced chemical equation of oxidation of oxalic acid by $\mathrm{KMnO}_4$ in acidic medium evolve 10 molecules of $\mathrm{CO}_2$.
AP EAMCET (Medical) 1998
Some Basic Concepts of Chemistry
228742
The volume in litres of $\mathrm{CO}_2$ liberated at STP when $10 \mathrm{~g}$ of $90 \%$ pure lime stone is heated completely is:
1 2.016
2 20.16
3 2.24
4 22.4
Explanation:
$\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\underset{\substack{22.4 \mathrm{LatatST} \\ \text { (1mole) }}}{\mathrm{CO}_2}$ $\therefore$ Molecular weight of $\mathrm{CaCO}_3=100 \mathrm{~g} / \mathrm{mole}$ Weight of $10 \mathrm{gm}$ of $90 \%$ pure limestone $=\frac{10 \times 90}{100}$ $=9$ gm pure $\mathrm{CaCO}_3$ $\therefore 9 \mathrm{gm}$ pure $\mathrm{CaCO}_3$ $\mathrm{CO}_2=\frac{22.4 \times 9}{100}=2.016 \mathrm{~L} \text {. }$
AP EAMCET (Medical) -1998 AP-EAMCET-1996
Some Basic Concepts of Chemistry
228743
The total number of valence electrons in $4.2 \mathrm{~g}$ of $\mathrm{N}_3^{-}$ion is ( $\mathrm{N}_{\mathrm{A}}$ is the Avogadro's number)
1 $2.1 \mathrm{~N}_{\mathrm{A}}$
2 $4.2 \mathrm{~N}_{\mathrm{A}}$
3 $1.6 \mathrm{~N}_{\mathrm{A}}$
4 $3.2 \mathrm{~N}_{\mathrm{A}}$
Explanation:
Molecular wt of $\mathrm{N}_3^{-}=3 \times 14=42 \mathrm{~g}$ Moles of $\mathrm{N}_3^{-}$ion $=\frac{4.2}{42}=0.1$ Each nitrogen atom has 5 valence electrons, total number of electrons in $\mathrm{N}_3^{-} \text {ion }=16$ Total number of electrons in 0.1 mole, $4.2 \mathrm{~g} \text { of } \mathrm{N}_3^{-} \text {ion }=0.1 \times 16 \times \mathrm{N}_{\mathrm{A}}=1.6 \mathrm{~N}_{\mathrm{A}}$
