228908
What is the value (in litres) of $\mathrm{CO}_2$ liberated at STP when $2.12 \mathrm{~g}$ of sodium carbonate (mol. wt. 106) is treated with excess dilute $\mathrm{HCl}$ ?
1 $2.28 \mathrm{~L}$
2 $0.448 \mathrm{~L}$
3 $44.8 \mathrm{~L}$
4 $22.4 \mathrm{~L}$
Explanation:
The balanced chemical equation is, $\begin{aligned} & \underset{\substack{106 \mathrm{~g} \\ 2.12 \mathrm{~g}}}{\mathrm{Na}_2 \mathrm{CO}_3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\underset{\substack{22.4 \mathrm{~L} \\ ?}}{\mathrm{CO}_2} \\ & \therefore \quad 106 \mathrm{gm} \mathrm{Na}_2 \mathrm{CO}_3 \text { required to } \mathrm{CO}_2 \text { is }=22.4 \mathrm{~L} \\ & \text { Volume of } \mathrm{CO}_2 \text { is required to } 2.12 \mathrm{gm} \text { of } \\ & \mathrm{Na}_2 \mathrm{CO}_3 \text { is }=\frac{22.4}{106} \times 2.12=0.448 \mathrm{~L} \\ \end{aligned}$ Thus, $2.12 \mathrm{~g}$ of sodium carbonate will produce $0.448 \mathrm{~L}$ at STP $\mathrm{CO}_2$.
AP-EAMCET-2000
Some Basic Concepts of Chemistry
228909
The molecular weight of an organic compound is 180. Its empirical formula is $\mathrm{CH}_2 \mathrm{O}$. Its molecular formula is
1 $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2 $\mathrm{C}_7 \mathrm{H}_{16} \mathrm{O}_5$
3 $\mathrm{C}_8 \mathrm{H}_4 \mathrm{O}_5$
4 $\mathrm{C}_5 \mathrm{H}_8 \mathrm{O}_7$
Explanation:
Given that, the empirical formula $=\mathrm{CH}_2 \mathrm{O}$ $\because$ Molecular weight of $\mathrm{CH}_2 \mathrm{O}=12+2+16=30$ $\Rightarrow$ No. of mole $(\mathrm{n})=\frac{180}{30}=6 \mathrm{~mole}$ Hence, the molecular formula $=\mathrm{n} \times$ empirical formula $=6 \times \mathrm{CH}_2 \mathrm{O}$ $=\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
A.P.EAMCET-1996
Some Basic Concepts of Chemistry
228910
The formula of a metal chloride is $\mathbf{M C l}_3$, it contains $20 \%$ of the metal. The atomic weight of the metal is approximately :
1 26.5
2 11.8
3 21.3
4 106.5
Explanation:
Given that $20 \mathrm{gm}$ of metal in $\mathrm{MCl}_3$. $\therefore \quad 80 \mathrm{gm}$ of $\mathrm{Cl}$ combines with $20 \mathrm{~g}$ of metal. $\therefore \quad 35.5 \mathrm{gm}$ of $\mathrm{Cl}$ combine with $=\frac{35.5 \times 20}{80}=8.875$ $\therefore$ Equivalent weight of metal $=8.875 \mathrm{~g}$. $\therefore$ Atomic weight $=$ Eq. wt. $\times$ valency $=8.875 \times 3$ $=26.5 \mathrm{gm}$.
AP-EAMCET-1994
Some Basic Concepts of Chemistry
228911
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
228908
What is the value (in litres) of $\mathrm{CO}_2$ liberated at STP when $2.12 \mathrm{~g}$ of sodium carbonate (mol. wt. 106) is treated with excess dilute $\mathrm{HCl}$ ?
1 $2.28 \mathrm{~L}$
2 $0.448 \mathrm{~L}$
3 $44.8 \mathrm{~L}$
4 $22.4 \mathrm{~L}$
Explanation:
The balanced chemical equation is, $\begin{aligned} & \underset{\substack{106 \mathrm{~g} \\ 2.12 \mathrm{~g}}}{\mathrm{Na}_2 \mathrm{CO}_3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\underset{\substack{22.4 \mathrm{~L} \\ ?}}{\mathrm{CO}_2} \\ & \therefore \quad 106 \mathrm{gm} \mathrm{Na}_2 \mathrm{CO}_3 \text { required to } \mathrm{CO}_2 \text { is }=22.4 \mathrm{~L} \\ & \text { Volume of } \mathrm{CO}_2 \text { is required to } 2.12 \mathrm{gm} \text { of } \\ & \mathrm{Na}_2 \mathrm{CO}_3 \text { is }=\frac{22.4}{106} \times 2.12=0.448 \mathrm{~L} \\ \end{aligned}$ Thus, $2.12 \mathrm{~g}$ of sodium carbonate will produce $0.448 \mathrm{~L}$ at STP $\mathrm{CO}_2$.
AP-EAMCET-2000
Some Basic Concepts of Chemistry
228909
The molecular weight of an organic compound is 180. Its empirical formula is $\mathrm{CH}_2 \mathrm{O}$. Its molecular formula is
1 $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2 $\mathrm{C}_7 \mathrm{H}_{16} \mathrm{O}_5$
3 $\mathrm{C}_8 \mathrm{H}_4 \mathrm{O}_5$
4 $\mathrm{C}_5 \mathrm{H}_8 \mathrm{O}_7$
Explanation:
Given that, the empirical formula $=\mathrm{CH}_2 \mathrm{O}$ $\because$ Molecular weight of $\mathrm{CH}_2 \mathrm{O}=12+2+16=30$ $\Rightarrow$ No. of mole $(\mathrm{n})=\frac{180}{30}=6 \mathrm{~mole}$ Hence, the molecular formula $=\mathrm{n} \times$ empirical formula $=6 \times \mathrm{CH}_2 \mathrm{O}$ $=\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
A.P.EAMCET-1996
Some Basic Concepts of Chemistry
228910
The formula of a metal chloride is $\mathbf{M C l}_3$, it contains $20 \%$ of the metal. The atomic weight of the metal is approximately :
1 26.5
2 11.8
3 21.3
4 106.5
Explanation:
Given that $20 \mathrm{gm}$ of metal in $\mathrm{MCl}_3$. $\therefore \quad 80 \mathrm{gm}$ of $\mathrm{Cl}$ combines with $20 \mathrm{~g}$ of metal. $\therefore \quad 35.5 \mathrm{gm}$ of $\mathrm{Cl}$ combine with $=\frac{35.5 \times 20}{80}=8.875$ $\therefore$ Equivalent weight of metal $=8.875 \mathrm{~g}$. $\therefore$ Atomic weight $=$ Eq. wt. $\times$ valency $=8.875 \times 3$ $=26.5 \mathrm{gm}$.
AP-EAMCET-1994
Some Basic Concepts of Chemistry
228911
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
228908
What is the value (in litres) of $\mathrm{CO}_2$ liberated at STP when $2.12 \mathrm{~g}$ of sodium carbonate (mol. wt. 106) is treated with excess dilute $\mathrm{HCl}$ ?
1 $2.28 \mathrm{~L}$
2 $0.448 \mathrm{~L}$
3 $44.8 \mathrm{~L}$
4 $22.4 \mathrm{~L}$
Explanation:
The balanced chemical equation is, $\begin{aligned} & \underset{\substack{106 \mathrm{~g} \\ 2.12 \mathrm{~g}}}{\mathrm{Na}_2 \mathrm{CO}_3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\underset{\substack{22.4 \mathrm{~L} \\ ?}}{\mathrm{CO}_2} \\ & \therefore \quad 106 \mathrm{gm} \mathrm{Na}_2 \mathrm{CO}_3 \text { required to } \mathrm{CO}_2 \text { is }=22.4 \mathrm{~L} \\ & \text { Volume of } \mathrm{CO}_2 \text { is required to } 2.12 \mathrm{gm} \text { of } \\ & \mathrm{Na}_2 \mathrm{CO}_3 \text { is }=\frac{22.4}{106} \times 2.12=0.448 \mathrm{~L} \\ \end{aligned}$ Thus, $2.12 \mathrm{~g}$ of sodium carbonate will produce $0.448 \mathrm{~L}$ at STP $\mathrm{CO}_2$.
AP-EAMCET-2000
Some Basic Concepts of Chemistry
228909
The molecular weight of an organic compound is 180. Its empirical formula is $\mathrm{CH}_2 \mathrm{O}$. Its molecular formula is
1 $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2 $\mathrm{C}_7 \mathrm{H}_{16} \mathrm{O}_5$
3 $\mathrm{C}_8 \mathrm{H}_4 \mathrm{O}_5$
4 $\mathrm{C}_5 \mathrm{H}_8 \mathrm{O}_7$
Explanation:
Given that, the empirical formula $=\mathrm{CH}_2 \mathrm{O}$ $\because$ Molecular weight of $\mathrm{CH}_2 \mathrm{O}=12+2+16=30$ $\Rightarrow$ No. of mole $(\mathrm{n})=\frac{180}{30}=6 \mathrm{~mole}$ Hence, the molecular formula $=\mathrm{n} \times$ empirical formula $=6 \times \mathrm{CH}_2 \mathrm{O}$ $=\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
A.P.EAMCET-1996
Some Basic Concepts of Chemistry
228910
The formula of a metal chloride is $\mathbf{M C l}_3$, it contains $20 \%$ of the metal. The atomic weight of the metal is approximately :
1 26.5
2 11.8
3 21.3
4 106.5
Explanation:
Given that $20 \mathrm{gm}$ of metal in $\mathrm{MCl}_3$. $\therefore \quad 80 \mathrm{gm}$ of $\mathrm{Cl}$ combines with $20 \mathrm{~g}$ of metal. $\therefore \quad 35.5 \mathrm{gm}$ of $\mathrm{Cl}$ combine with $=\frac{35.5 \times 20}{80}=8.875$ $\therefore$ Equivalent weight of metal $=8.875 \mathrm{~g}$. $\therefore$ Atomic weight $=$ Eq. wt. $\times$ valency $=8.875 \times 3$ $=26.5 \mathrm{gm}$.
AP-EAMCET-1994
Some Basic Concepts of Chemistry
228911
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
228908
What is the value (in litres) of $\mathrm{CO}_2$ liberated at STP when $2.12 \mathrm{~g}$ of sodium carbonate (mol. wt. 106) is treated with excess dilute $\mathrm{HCl}$ ?
1 $2.28 \mathrm{~L}$
2 $0.448 \mathrm{~L}$
3 $44.8 \mathrm{~L}$
4 $22.4 \mathrm{~L}$
Explanation:
The balanced chemical equation is, $\begin{aligned} & \underset{\substack{106 \mathrm{~g} \\ 2.12 \mathrm{~g}}}{\mathrm{Na}_2 \mathrm{CO}_3}+2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}+\underset{\substack{22.4 \mathrm{~L} \\ ?}}{\mathrm{CO}_2} \\ & \therefore \quad 106 \mathrm{gm} \mathrm{Na}_2 \mathrm{CO}_3 \text { required to } \mathrm{CO}_2 \text { is }=22.4 \mathrm{~L} \\ & \text { Volume of } \mathrm{CO}_2 \text { is required to } 2.12 \mathrm{gm} \text { of } \\ & \mathrm{Na}_2 \mathrm{CO}_3 \text { is }=\frac{22.4}{106} \times 2.12=0.448 \mathrm{~L} \\ \end{aligned}$ Thus, $2.12 \mathrm{~g}$ of sodium carbonate will produce $0.448 \mathrm{~L}$ at STP $\mathrm{CO}_2$.
AP-EAMCET-2000
Some Basic Concepts of Chemistry
228909
The molecular weight of an organic compound is 180. Its empirical formula is $\mathrm{CH}_2 \mathrm{O}$. Its molecular formula is
1 $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
2 $\mathrm{C}_7 \mathrm{H}_{16} \mathrm{O}_5$
3 $\mathrm{C}_8 \mathrm{H}_4 \mathrm{O}_5$
4 $\mathrm{C}_5 \mathrm{H}_8 \mathrm{O}_7$
Explanation:
Given that, the empirical formula $=\mathrm{CH}_2 \mathrm{O}$ $\because$ Molecular weight of $\mathrm{CH}_2 \mathrm{O}=12+2+16=30$ $\Rightarrow$ No. of mole $(\mathrm{n})=\frac{180}{30}=6 \mathrm{~mole}$ Hence, the molecular formula $=\mathrm{n} \times$ empirical formula $=6 \times \mathrm{CH}_2 \mathrm{O}$ $=\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
A.P.EAMCET-1996
Some Basic Concepts of Chemistry
228910
The formula of a metal chloride is $\mathbf{M C l}_3$, it contains $20 \%$ of the metal. The atomic weight of the metal is approximately :
1 26.5
2 11.8
3 21.3
4 106.5
Explanation:
Given that $20 \mathrm{gm}$ of metal in $\mathrm{MCl}_3$. $\therefore \quad 80 \mathrm{gm}$ of $\mathrm{Cl}$ combines with $20 \mathrm{~g}$ of metal. $\therefore \quad 35.5 \mathrm{gm}$ of $\mathrm{Cl}$ combine with $=\frac{35.5 \times 20}{80}=8.875$ $\therefore$ Equivalent weight of metal $=8.875 \mathrm{~g}$. $\therefore$ Atomic weight $=$ Eq. wt. $\times$ valency $=8.875 \times 3$ $=26.5 \mathrm{gm}$.
AP-EAMCET-1994
Some Basic Concepts of Chemistry
228911
The number of molecules present in $3.5 \mathrm{~g}$ of $\mathrm{CO}$ at $0^{\circ} \mathrm{C}$ and $760 \mathrm{~mm}$ pressure is : 7. Empirical Formula for
1 $6.02 \times 10^{23}$
2 $1.25 \times 6.02 \times 10^{23}$
3 $0.125 \times 6.02 \times 10^{23}$
4 $1.25 \mathrm{~N}_{\mathrm{A}}$
Explanation:
28 gm CO at STP $=6.023 \times 10^{23}$ molecules $\therefore \quad 3.5 \mathrm{gm}$ of CO at STP $=\frac{1}{28} \times 3.5 \times 6.023 \times 10^{23}$ $=0.125 \times 6.02 \times 10^{23}$ Molecular Formula
