228902
$64 \mathrm{~g}$ of an organic compound contains $24 \mathrm{~g}$ of carbon, 8 g of hydrogen and the rest oxygen. The empirical formula of the compound is
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{C}_2 \mathrm{H}_8 \mathrm{O}_2$
Explanation:
Mass of oxygen $=64-24-8=32$ | Element | Mole ratio | Simple ratio | | :--- | :---: | :---: | | C | $\frac{24}{12}=2$ | $\frac{2}{2}=1$ | | H | $\frac{8}{1}=8$ | $\frac{8}{2}=4$ | | $\mathrm{O}$ | $\frac{32}{16}=2$ | $\frac{2}{2}=1$ | Empirical formula is $\mathrm{CH}_4 \mathrm{O}$.
CG PET-2007
Some Basic Concepts of Chemistry
228903
What is the stoichiometric coefficient of $\mathrm{Ca}$ in the reaction? $\mathrm{Ca}+\mathrm{Al}^{3+} \rightarrow \mathrm{Ca}^{2+}+\mathrm{Al}$
1 2
2 1
3 3
4 4
Explanation:
$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}$ $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}$ Eq. (i) multiply by 3 and Eq. (ii) multiply by 2 , and add Eqs. (i) and (ii) $\begin{aligned} & 3 \mathrm{Ca} \rightarrow 3 \mathrm{Ca}^{2+}+6 \mathrm{e}^{-} \\ & 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \\ & 3 \mathrm{Ca}+2 \mathrm{Al}^{3+} \rightarrow 3 \mathrm{Ca}^{2+}+2 \mathrm{Al} \end{aligned}$ Therefore, the stoichiometric coefficient of $\mathrm{Ca}$ in the given reaction is 3 .
UP CPMT-2007
Some Basic Concepts of Chemistry
228904
Value of $x$ in potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_x\left(\mathrm{SO}_4\right)_3$ $\cdot 2 \mathrm{HH}_2 \mathrm{O}$ is
1 4
2 1
3 2
4 None of these
Explanation:
Potash alum is double salt. Given Potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_{\mathrm{x}}\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ But formula of potash alum is $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ After comparing both equations, Value of $x=2$
UP CPMT-2007
Some Basic Concepts of Chemistry
228905
The number of molecules of $\mathrm{CO}_2$ present in $44 \mathrm{~g}$ of $\mathrm{CO}_2$ is :
1 $6.0 \times 10^{23}$
2 $3 \times 10^{23}$
3 $12 \times 10^{23}$
4 $3 \times 10^{10}$
Explanation:
Given mass of $\mathrm{CO}_2, \mathrm{~W}=44 \mathrm{~g}$ Molar mass of $\mathrm{CO}_2, \mathrm{M}=44 \mathrm{~g} / \mathrm{mol}$ Number of moles, $\mathrm{n}=\frac{\mathrm{W}}{\mathrm{M}}=\frac{44}{44}=1 \mathrm{~mol}$ Number of molecules, $\mathrm{N}=\mathrm{n} \times \mathrm{N}_{\mathrm{A}}$ $=1 \times 6.022 \times 10^{23}$ $=6.022 \times 10^{23} \text { molecule. }$
228902
$64 \mathrm{~g}$ of an organic compound contains $24 \mathrm{~g}$ of carbon, 8 g of hydrogen and the rest oxygen. The empirical formula of the compound is
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{C}_2 \mathrm{H}_8 \mathrm{O}_2$
Explanation:
Mass of oxygen $=64-24-8=32$ | Element | Mole ratio | Simple ratio | | :--- | :---: | :---: | | C | $\frac{24}{12}=2$ | $\frac{2}{2}=1$ | | H | $\frac{8}{1}=8$ | $\frac{8}{2}=4$ | | $\mathrm{O}$ | $\frac{32}{16}=2$ | $\frac{2}{2}=1$ | Empirical formula is $\mathrm{CH}_4 \mathrm{O}$.
CG PET-2007
Some Basic Concepts of Chemistry
228903
What is the stoichiometric coefficient of $\mathrm{Ca}$ in the reaction? $\mathrm{Ca}+\mathrm{Al}^{3+} \rightarrow \mathrm{Ca}^{2+}+\mathrm{Al}$
1 2
2 1
3 3
4 4
Explanation:
$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}$ $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}$ Eq. (i) multiply by 3 and Eq. (ii) multiply by 2 , and add Eqs. (i) and (ii) $\begin{aligned} & 3 \mathrm{Ca} \rightarrow 3 \mathrm{Ca}^{2+}+6 \mathrm{e}^{-} \\ & 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \\ & 3 \mathrm{Ca}+2 \mathrm{Al}^{3+} \rightarrow 3 \mathrm{Ca}^{2+}+2 \mathrm{Al} \end{aligned}$ Therefore, the stoichiometric coefficient of $\mathrm{Ca}$ in the given reaction is 3 .
UP CPMT-2007
Some Basic Concepts of Chemistry
228904
Value of $x$ in potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_x\left(\mathrm{SO}_4\right)_3$ $\cdot 2 \mathrm{HH}_2 \mathrm{O}$ is
1 4
2 1
3 2
4 None of these
Explanation:
Potash alum is double salt. Given Potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_{\mathrm{x}}\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ But formula of potash alum is $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ After comparing both equations, Value of $x=2$
UP CPMT-2007
Some Basic Concepts of Chemistry
228905
The number of molecules of $\mathrm{CO}_2$ present in $44 \mathrm{~g}$ of $\mathrm{CO}_2$ is :
1 $6.0 \times 10^{23}$
2 $3 \times 10^{23}$
3 $12 \times 10^{23}$
4 $3 \times 10^{10}$
Explanation:
Given mass of $\mathrm{CO}_2, \mathrm{~W}=44 \mathrm{~g}$ Molar mass of $\mathrm{CO}_2, \mathrm{M}=44 \mathrm{~g} / \mathrm{mol}$ Number of moles, $\mathrm{n}=\frac{\mathrm{W}}{\mathrm{M}}=\frac{44}{44}=1 \mathrm{~mol}$ Number of molecules, $\mathrm{N}=\mathrm{n} \times \mathrm{N}_{\mathrm{A}}$ $=1 \times 6.022 \times 10^{23}$ $=6.022 \times 10^{23} \text { molecule. }$
228902
$64 \mathrm{~g}$ of an organic compound contains $24 \mathrm{~g}$ of carbon, 8 g of hydrogen and the rest oxygen. The empirical formula of the compound is
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{C}_2 \mathrm{H}_8 \mathrm{O}_2$
Explanation:
Mass of oxygen $=64-24-8=32$ | Element | Mole ratio | Simple ratio | | :--- | :---: | :---: | | C | $\frac{24}{12}=2$ | $\frac{2}{2}=1$ | | H | $\frac{8}{1}=8$ | $\frac{8}{2}=4$ | | $\mathrm{O}$ | $\frac{32}{16}=2$ | $\frac{2}{2}=1$ | Empirical formula is $\mathrm{CH}_4 \mathrm{O}$.
CG PET-2007
Some Basic Concepts of Chemistry
228903
What is the stoichiometric coefficient of $\mathrm{Ca}$ in the reaction? $\mathrm{Ca}+\mathrm{Al}^{3+} \rightarrow \mathrm{Ca}^{2+}+\mathrm{Al}$
1 2
2 1
3 3
4 4
Explanation:
$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}$ $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}$ Eq. (i) multiply by 3 and Eq. (ii) multiply by 2 , and add Eqs. (i) and (ii) $\begin{aligned} & 3 \mathrm{Ca} \rightarrow 3 \mathrm{Ca}^{2+}+6 \mathrm{e}^{-} \\ & 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \\ & 3 \mathrm{Ca}+2 \mathrm{Al}^{3+} \rightarrow 3 \mathrm{Ca}^{2+}+2 \mathrm{Al} \end{aligned}$ Therefore, the stoichiometric coefficient of $\mathrm{Ca}$ in the given reaction is 3 .
UP CPMT-2007
Some Basic Concepts of Chemistry
228904
Value of $x$ in potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_x\left(\mathrm{SO}_4\right)_3$ $\cdot 2 \mathrm{HH}_2 \mathrm{O}$ is
1 4
2 1
3 2
4 None of these
Explanation:
Potash alum is double salt. Given Potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_{\mathrm{x}}\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ But formula of potash alum is $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ After comparing both equations, Value of $x=2$
UP CPMT-2007
Some Basic Concepts of Chemistry
228905
The number of molecules of $\mathrm{CO}_2$ present in $44 \mathrm{~g}$ of $\mathrm{CO}_2$ is :
1 $6.0 \times 10^{23}$
2 $3 \times 10^{23}$
3 $12 \times 10^{23}$
4 $3 \times 10^{10}$
Explanation:
Given mass of $\mathrm{CO}_2, \mathrm{~W}=44 \mathrm{~g}$ Molar mass of $\mathrm{CO}_2, \mathrm{M}=44 \mathrm{~g} / \mathrm{mol}$ Number of moles, $\mathrm{n}=\frac{\mathrm{W}}{\mathrm{M}}=\frac{44}{44}=1 \mathrm{~mol}$ Number of molecules, $\mathrm{N}=\mathrm{n} \times \mathrm{N}_{\mathrm{A}}$ $=1 \times 6.022 \times 10^{23}$ $=6.022 \times 10^{23} \text { molecule. }$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Some Basic Concepts of Chemistry
228902
$64 \mathrm{~g}$ of an organic compound contains $24 \mathrm{~g}$ of carbon, 8 g of hydrogen and the rest oxygen. The empirical formula of the compound is
1 $\mathrm{CH}_2 \mathrm{O}$
2 $\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}$
3 $\mathrm{CH}_4 \mathrm{O}$
4 $\mathrm{C}_2 \mathrm{H}_8 \mathrm{O}_2$
Explanation:
Mass of oxygen $=64-24-8=32$ | Element | Mole ratio | Simple ratio | | :--- | :---: | :---: | | C | $\frac{24}{12}=2$ | $\frac{2}{2}=1$ | | H | $\frac{8}{1}=8$ | $\frac{8}{2}=4$ | | $\mathrm{O}$ | $\frac{32}{16}=2$ | $\frac{2}{2}=1$ | Empirical formula is $\mathrm{CH}_4 \mathrm{O}$.
CG PET-2007
Some Basic Concepts of Chemistry
228903
What is the stoichiometric coefficient of $\mathrm{Ca}$ in the reaction? $\mathrm{Ca}+\mathrm{Al}^{3+} \rightarrow \mathrm{Ca}^{2+}+\mathrm{Al}$
1 2
2 1
3 3
4 4
Explanation:
$\mathrm{Ca} \rightarrow \mathrm{Ca}^{2+}+2 \mathrm{e}^{-}$ $\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}$ Eq. (i) multiply by 3 and Eq. (ii) multiply by 2 , and add Eqs. (i) and (ii) $\begin{aligned} & 3 \mathrm{Ca} \rightarrow 3 \mathrm{Ca}^{2+}+6 \mathrm{e}^{-} \\ & 2 \mathrm{Al}^{3+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Al} \\ & 3 \mathrm{Ca}+2 \mathrm{Al}^{3+} \rightarrow 3 \mathrm{Ca}^{2+}+2 \mathrm{Al} \end{aligned}$ Therefore, the stoichiometric coefficient of $\mathrm{Ca}$ in the given reaction is 3 .
UP CPMT-2007
Some Basic Concepts of Chemistry
228904
Value of $x$ in potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_x\left(\mathrm{SO}_4\right)_3$ $\cdot 2 \mathrm{HH}_2 \mathrm{O}$ is
1 4
2 1
3 2
4 None of these
Explanation:
Potash alum is double salt. Given Potash alum, $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_{\mathrm{x}}\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ But formula of potash alum is $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$ After comparing both equations, Value of $x=2$
UP CPMT-2007
Some Basic Concepts of Chemistry
228905
The number of molecules of $\mathrm{CO}_2$ present in $44 \mathrm{~g}$ of $\mathrm{CO}_2$ is :
1 $6.0 \times 10^{23}$
2 $3 \times 10^{23}$
3 $12 \times 10^{23}$
4 $3 \times 10^{10}$
Explanation:
Given mass of $\mathrm{CO}_2, \mathrm{~W}=44 \mathrm{~g}$ Molar mass of $\mathrm{CO}_2, \mathrm{M}=44 \mathrm{~g} / \mathrm{mol}$ Number of moles, $\mathrm{n}=\frac{\mathrm{W}}{\mathrm{M}}=\frac{44}{44}=1 \mathrm{~mol}$ Number of molecules, $\mathrm{N}=\mathrm{n} \times \mathrm{N}_{\mathrm{A}}$ $=1 \times 6.022 \times 10^{23}$ $=6.022 \times 10^{23} \text { molecule. }$
