228877
In a non-stoichiometric sample of cuprous sulphide. With the composition $\mathrm{Cu}_{1.8} \mathrm{~S}$. Cupric ions are also present in the lattice. What mole percent of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal?
1 $99.99 \%$
2 $11.11 \%$
3 $88.88 \%$
4 $18 \%$ AP
Explanation:
Exp : Given that, Non-Stoichiometric sample of cuprous sulphide $\mathrm{Cu}_{1.8} \mathrm{~S}-\left[\begin{array}{l} 10 \times 2=20(-\mathrm{ve}) \text { charge } \\ 18 \times 1=18(+\mathrm{ve}) \text { charge } \end{array}\right.$ $\begin{aligned} & \Rightarrow 2 \mathrm{Cu}^{2+} \text { ions }=\frac{2}{18} \times 100=\frac{1}{\mathrm{a}} \times 100 \\ & =11.11 \% \\ & =16 \mathrm{Cu}^{+} \text {ions }=\frac{16}{18} \times 100 \\ & =\frac{8}{4} \times 100=88.89 \% \\ & \end{aligned}$ Hence, $11.11 \%$ of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal.
Shift-II
Some Basic Concepts of Chemistry
228878
For the reaction: $\mathrm{NH}_3+\mathrm{OCl}^{-} \longrightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}$ in basic medium, the coefficients of $\mathrm{NH}_3, \mathrm{OCl}^{-}$ and $\mathrm{N}_2 \mathrm{H}_4$ for the balanced equation are respectively
1 2,2,2
2 $2,2,1$
3 $2,1,1$
4 $4,4,2$
Explanation:
For ammonia $2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 \mathrm{H}_4+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \text {(Basic medium) }$ For $\mathrm{OCl}^{-}$ $\mathrm{OCl}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Adding both equation (i) and (ii) $2 \mathrm{NH}_3+\mathrm{OCl}^{-} \rightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Co-efficients of $\mathrm{NH}_3, \mathrm{ClO}^{-}$and $\mathrm{N}_2 \mathrm{H}_4$ are 2,1 and 1 .
BITSAT-2021
Some Basic Concepts of Chemistry
228879
A pure compound contains $2.4 \mathrm{~g}$ of $\mathrm{C}, 1.2 \times 10^{-23}$ atoms of $\mathrm{H}, 0.2$ moles of oxygen atoms. Its empirical formula is
1 $\mathrm{C}_2 \mathrm{HO}$
2 $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2$
3 $\mathrm{CH}_2 \mathrm{O}$
4 $\mathrm{CHO}$
Explanation:
Convert each element into number of moles. Moles of carbon $=\frac{2.4}{12}=0.2$ Moles of hydrogen $=\frac{1.2 \times 10^{23}}{6.023 \times 10^{23}}=0.199$ Moles of oxygen $=0.2$ Empirical formula is $\mathrm{CHO}$.
228877
In a non-stoichiometric sample of cuprous sulphide. With the composition $\mathrm{Cu}_{1.8} \mathrm{~S}$. Cupric ions are also present in the lattice. What mole percent of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal?
1 $99.99 \%$
2 $11.11 \%$
3 $88.88 \%$
4 $18 \%$ AP
Explanation:
Exp : Given that, Non-Stoichiometric sample of cuprous sulphide $\mathrm{Cu}_{1.8} \mathrm{~S}-\left[\begin{array}{l} 10 \times 2=20(-\mathrm{ve}) \text { charge } \\ 18 \times 1=18(+\mathrm{ve}) \text { charge } \end{array}\right.$ $\begin{aligned} & \Rightarrow 2 \mathrm{Cu}^{2+} \text { ions }=\frac{2}{18} \times 100=\frac{1}{\mathrm{a}} \times 100 \\ & =11.11 \% \\ & =16 \mathrm{Cu}^{+} \text {ions }=\frac{16}{18} \times 100 \\ & =\frac{8}{4} \times 100=88.89 \% \\ & \end{aligned}$ Hence, $11.11 \%$ of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal.
Shift-II
Some Basic Concepts of Chemistry
228878
For the reaction: $\mathrm{NH}_3+\mathrm{OCl}^{-} \longrightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}$ in basic medium, the coefficients of $\mathrm{NH}_3, \mathrm{OCl}^{-}$ and $\mathrm{N}_2 \mathrm{H}_4$ for the balanced equation are respectively
1 2,2,2
2 $2,2,1$
3 $2,1,1$
4 $4,4,2$
Explanation:
For ammonia $2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 \mathrm{H}_4+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \text {(Basic medium) }$ For $\mathrm{OCl}^{-}$ $\mathrm{OCl}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Adding both equation (i) and (ii) $2 \mathrm{NH}_3+\mathrm{OCl}^{-} \rightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Co-efficients of $\mathrm{NH}_3, \mathrm{ClO}^{-}$and $\mathrm{N}_2 \mathrm{H}_4$ are 2,1 and 1 .
BITSAT-2021
Some Basic Concepts of Chemistry
228879
A pure compound contains $2.4 \mathrm{~g}$ of $\mathrm{C}, 1.2 \times 10^{-23}$ atoms of $\mathrm{H}, 0.2$ moles of oxygen atoms. Its empirical formula is
1 $\mathrm{C}_2 \mathrm{HO}$
2 $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2$
3 $\mathrm{CH}_2 \mathrm{O}$
4 $\mathrm{CHO}$
Explanation:
Convert each element into number of moles. Moles of carbon $=\frac{2.4}{12}=0.2$ Moles of hydrogen $=\frac{1.2 \times 10^{23}}{6.023 \times 10^{23}}=0.199$ Moles of oxygen $=0.2$ Empirical formula is $\mathrm{CHO}$.
228877
In a non-stoichiometric sample of cuprous sulphide. With the composition $\mathrm{Cu}_{1.8} \mathrm{~S}$. Cupric ions are also present in the lattice. What mole percent of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal?
1 $99.99 \%$
2 $11.11 \%$
3 $88.88 \%$
4 $18 \%$ AP
Explanation:
Exp : Given that, Non-Stoichiometric sample of cuprous sulphide $\mathrm{Cu}_{1.8} \mathrm{~S}-\left[\begin{array}{l} 10 \times 2=20(-\mathrm{ve}) \text { charge } \\ 18 \times 1=18(+\mathrm{ve}) \text { charge } \end{array}\right.$ $\begin{aligned} & \Rightarrow 2 \mathrm{Cu}^{2+} \text { ions }=\frac{2}{18} \times 100=\frac{1}{\mathrm{a}} \times 100 \\ & =11.11 \% \\ & =16 \mathrm{Cu}^{+} \text {ions }=\frac{16}{18} \times 100 \\ & =\frac{8}{4} \times 100=88.89 \% \\ & \end{aligned}$ Hence, $11.11 \%$ of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal.
Shift-II
Some Basic Concepts of Chemistry
228878
For the reaction: $\mathrm{NH}_3+\mathrm{OCl}^{-} \longrightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}$ in basic medium, the coefficients of $\mathrm{NH}_3, \mathrm{OCl}^{-}$ and $\mathrm{N}_2 \mathrm{H}_4$ for the balanced equation are respectively
1 2,2,2
2 $2,2,1$
3 $2,1,1$
4 $4,4,2$
Explanation:
For ammonia $2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 \mathrm{H}_4+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \text {(Basic medium) }$ For $\mathrm{OCl}^{-}$ $\mathrm{OCl}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Adding both equation (i) and (ii) $2 \mathrm{NH}_3+\mathrm{OCl}^{-} \rightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Co-efficients of $\mathrm{NH}_3, \mathrm{ClO}^{-}$and $\mathrm{N}_2 \mathrm{H}_4$ are 2,1 and 1 .
BITSAT-2021
Some Basic Concepts of Chemistry
228879
A pure compound contains $2.4 \mathrm{~g}$ of $\mathrm{C}, 1.2 \times 10^{-23}$ atoms of $\mathrm{H}, 0.2$ moles of oxygen atoms. Its empirical formula is
1 $\mathrm{C}_2 \mathrm{HO}$
2 $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2$
3 $\mathrm{CH}_2 \mathrm{O}$
4 $\mathrm{CHO}$
Explanation:
Convert each element into number of moles. Moles of carbon $=\frac{2.4}{12}=0.2$ Moles of hydrogen $=\frac{1.2 \times 10^{23}}{6.023 \times 10^{23}}=0.199$ Moles of oxygen $=0.2$ Empirical formula is $\mathrm{CHO}$.
228877
In a non-stoichiometric sample of cuprous sulphide. With the composition $\mathrm{Cu}_{1.8} \mathrm{~S}$. Cupric ions are also present in the lattice. What mole percent of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal?
1 $99.99 \%$
2 $11.11 \%$
3 $88.88 \%$
4 $18 \%$ AP
Explanation:
Exp : Given that, Non-Stoichiometric sample of cuprous sulphide $\mathrm{Cu}_{1.8} \mathrm{~S}-\left[\begin{array}{l} 10 \times 2=20(-\mathrm{ve}) \text { charge } \\ 18 \times 1=18(+\mathrm{ve}) \text { charge } \end{array}\right.$ $\begin{aligned} & \Rightarrow 2 \mathrm{Cu}^{2+} \text { ions }=\frac{2}{18} \times 100=\frac{1}{\mathrm{a}} \times 100 \\ & =11.11 \% \\ & =16 \mathrm{Cu}^{+} \text {ions }=\frac{16}{18} \times 100 \\ & =\frac{8}{4} \times 100=88.89 \% \\ & \end{aligned}$ Hence, $11.11 \%$ of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal.
Shift-II
Some Basic Concepts of Chemistry
228878
For the reaction: $\mathrm{NH}_3+\mathrm{OCl}^{-} \longrightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}$ in basic medium, the coefficients of $\mathrm{NH}_3, \mathrm{OCl}^{-}$ and $\mathrm{N}_2 \mathrm{H}_4$ for the balanced equation are respectively
1 2,2,2
2 $2,2,1$
3 $2,1,1$
4 $4,4,2$
Explanation:
For ammonia $2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 \mathrm{H}_4+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \text {(Basic medium) }$ For $\mathrm{OCl}^{-}$ $\mathrm{OCl}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Adding both equation (i) and (ii) $2 \mathrm{NH}_3+\mathrm{OCl}^{-} \rightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Co-efficients of $\mathrm{NH}_3, \mathrm{ClO}^{-}$and $\mathrm{N}_2 \mathrm{H}_4$ are 2,1 and 1 .
BITSAT-2021
Some Basic Concepts of Chemistry
228879
A pure compound contains $2.4 \mathrm{~g}$ of $\mathrm{C}, 1.2 \times 10^{-23}$ atoms of $\mathrm{H}, 0.2$ moles of oxygen atoms. Its empirical formula is
1 $\mathrm{C}_2 \mathrm{HO}$
2 $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2$
3 $\mathrm{CH}_2 \mathrm{O}$
4 $\mathrm{CHO}$
Explanation:
Convert each element into number of moles. Moles of carbon $=\frac{2.4}{12}=0.2$ Moles of hydrogen $=\frac{1.2 \times 10^{23}}{6.023 \times 10^{23}}=0.199$ Moles of oxygen $=0.2$ Empirical formula is $\mathrm{CHO}$.
228877
In a non-stoichiometric sample of cuprous sulphide. With the composition $\mathrm{Cu}_{1.8} \mathrm{~S}$. Cupric ions are also present in the lattice. What mole percent of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal?
1 $99.99 \%$
2 $11.11 \%$
3 $88.88 \%$
4 $18 \%$ AP
Explanation:
Exp : Given that, Non-Stoichiometric sample of cuprous sulphide $\mathrm{Cu}_{1.8} \mathrm{~S}-\left[\begin{array}{l} 10 \times 2=20(-\mathrm{ve}) \text { charge } \\ 18 \times 1=18(+\mathrm{ve}) \text { charge } \end{array}\right.$ $\begin{aligned} & \Rightarrow 2 \mathrm{Cu}^{2+} \text { ions }=\frac{2}{18} \times 100=\frac{1}{\mathrm{a}} \times 100 \\ & =11.11 \% \\ & =16 \mathrm{Cu}^{+} \text {ions }=\frac{16}{18} \times 100 \\ & =\frac{8}{4} \times 100=88.89 \% \\ & \end{aligned}$ Hence, $11.11 \%$ of $\mathrm{Cu}^{2+}$ is present in the copper content of the crystal.
Shift-II
Some Basic Concepts of Chemistry
228878
For the reaction: $\mathrm{NH}_3+\mathrm{OCl}^{-} \longrightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}$ in basic medium, the coefficients of $\mathrm{NH}_3, \mathrm{OCl}^{-}$ and $\mathrm{N}_2 \mathrm{H}_4$ for the balanced equation are respectively
1 2,2,2
2 $2,2,1$
3 $2,1,1$
4 $4,4,2$
Explanation:
For ammonia $2 \mathrm{NH}_3 \rightarrow \mathrm{N}_2 \mathrm{H}_4+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \text {(Basic medium) }$ For $\mathrm{OCl}^{-}$ $\mathrm{OCl}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Adding both equation (i) and (ii) $2 \mathrm{NH}_3+\mathrm{OCl}^{-} \rightarrow \mathrm{N}_2 \mathrm{H}_4+\mathrm{Cl}^{-}+\mathrm{H}_2 \mathrm{O}$ Co-efficients of $\mathrm{NH}_3, \mathrm{ClO}^{-}$and $\mathrm{N}_2 \mathrm{H}_4$ are 2,1 and 1 .
BITSAT-2021
Some Basic Concepts of Chemistry
228879
A pure compound contains $2.4 \mathrm{~g}$ of $\mathrm{C}, 1.2 \times 10^{-23}$ atoms of $\mathrm{H}, 0.2$ moles of oxygen atoms. Its empirical formula is
1 $\mathrm{C}_2 \mathrm{HO}$
2 $\mathrm{C}_2 \mathrm{H}_2 \mathrm{O}_2$
3 $\mathrm{CH}_2 \mathrm{O}$
4 $\mathrm{CHO}$
Explanation:
Convert each element into number of moles. Moles of carbon $=\frac{2.4}{12}=0.2$ Moles of hydrogen $=\frac{1.2 \times 10^{23}}{6.023 \times 10^{23}}=0.199$ Moles of oxygen $=0.2$ Empirical formula is $\mathrm{CHO}$.
