Difficult
$0.45 \ g$ of an organic compound on combustion gives $0.792 \ g$ of $CO_2$ and $0.324 \ g$ of $H_2O$. In the Kjeldahl method,$0.24 \ g$ of the same compound yields ammonia,which is absorbed in $50 \ mL$ of $0.25 \ N \ H_2SO_4$. The excess acid requires $77.0 \ mL$ of $0.25 \ N \ NaOH$ for neutralization. Determine the empirical formula of the compound. (Atomic masses: $C = 12, H = 1, O = 16, N = 14, S = 32$)
(D) $1$. Calculate percentage of $C$: $\text{Mass of } C = (12/44) \times 0.792 = 0.216 \ g$. $\% C = (0.216 / 0.45) \times 100 = 48\%$.
$2$. Calculate percentage of $H$: $\text{Mass of } H = (2/18) \times 0.324 = 0.036 \ g$. $\% H = (0.036 / 0.45) \times 100 = 8\%$.
$3$. Calculate percentage of $N$: $\text{Total } H_2SO_4 = 50 \times 0.25 = 12.5 \ \text{meq}$. $\text{Acid used for } NH_3 = 12.5 - (77.0 \times 0.25 / 2) = 12.5 - 9.625 = 2.875 \ \text{meq}$. $\text{Mass of } N = (2.875 \times 14) / 1000 = 0.04025 \ g$. $\% N = (0.04025 / 0.24) \times 100 \approx 16.77\%$.
$4$. Calculate percentage of $O$: $\% O = 100 - (48 + 8 + 16.77) = 27.23\%$.
$5$. Empirical formula calculation: $C: 48/12 = 4, H: 8/1 = 8, N: 16.77/14 \approx 1.2, O: 27.23/16 \approx 1.7$. Dividing by smallest $(1.2)$: $C: 3.33, H: 6.66, N: 1, O: 1.4$. Multiplying by $3$: $C: 10, H: 20, N: 3, O: 4$. The empirical formula is $C_{10}H_{20}N_3O_4$.
$2$. Calculate percentage of $H$: $\text{Mass of } H = (2/18) \times 0.324 = 0.036 \ g$. $\% H = (0.036 / 0.45) \times 100 = 8\%$.
$3$. Calculate percentage of $N$: $\text{Total } H_2SO_4 = 50 \times 0.25 = 12.5 \ \text{meq}$. $\text{Acid used for } NH_3 = 12.5 - (77.0 \times 0.25 / 2) = 12.5 - 9.625 = 2.875 \ \text{meq}$. $\text{Mass of } N = (2.875 \times 14) / 1000 = 0.04025 \ g$. $\% N = (0.04025 / 0.24) \times 100 \approx 16.77\%$.
$4$. Calculate percentage of $O$: $\% O = 100 - (48 + 8 + 16.77) = 27.23\%$.
$5$. Empirical formula calculation: $C: 48/12 = 4, H: 8/1 = 8, N: 16.77/14 \approx 1.2, O: 27.23/16 \approx 1.7$. Dividing by smallest $(1.2)$: $C: 3.33, H: 6.66, N: 1, O: 1.4$. Multiplying by $3$: $C: 10, H: 20, N: 3, O: 4$. The empirical formula is $C_{10}H_{20}N_3O_4$.
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Which of the following is $NOT$ true for $1 \ g$ of Helium?
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Column-$I$
Column-$II$
$A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$
$I$. $32 \ g$
$B$. $1.2 \times 10^{24}$ atoms of oxygen
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$C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$
$III$. $42 \ g$
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$IV$. $16 \ g$
The correct match is:
| Column-$I$ | Column-$II$ |
|---|---|
| $A$. $5.6 \ L$ of $SO_{2(g)}$ at $STP$ | $I$. $32 \ g$ |
| $B$. $1.2 \times 10^{24}$ atoms of oxygen | $II$. $66 \ g$ |
| $C$. $33.6 \ L$ of $CO_{2(g)}$ at $STP$ | $III$. $42 \ g$ |
| $D$. $1.5 \ g$-molecule of $N_2$ | $IV$. $16 \ g$ |
The correct match is:
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$(i)$ The molarity of a $0.6 \ N$ aqueous solution of ${H_3PO_3}$ is $1.2 \ M$.
$(ii)$ $\text{Empirical formula} = n \times \text{Molecular formula}$.
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$(i)$ The molarity of a $0.6 \ N$ aqueous solution of ${H_3PO_3}$ is $1.2 \ M$.
$(ii)$ $\text{Empirical formula} = n \times \text{Molecular formula}$.
$(iii)$ In ${Cl_2O}$,the weight of chlorine combining with $8 \ g$ of oxygen is $35.5 \ g$.
Medium
View SolutionSubstance $A_2B_{(g)}$ can undergo decomposition to form two sets of products. If the molar ratio of $A_{2(g)}$ to $A_{(g)}$ is $5 : 3$ in a set of product gases,then the energy involved in the decomposition of $1 \, \text{mole}$ of $A_2B_{(g)}$ is:
$A_2B_{(g)} \to A_{2(g)} + B_{(g)}; \Delta H^o = 40 \, \text{kJ/mol}$
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$A_2B_{(g)} \to A_{2(g)} + B_{(g)}; \Delta H^o = 40 \, \text{kJ/mol}$
$A_2B_{(g)} \to A_{(g)} + AB_{(g)}; \Delta H^o = 50 \, \text{kJ/mol}$
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View SolutionThe ratio of the number of water molecules in Mohr's salt and potash alum is $.... \times 10^{-1}$.
DifficultJEE MAIN 2021
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