An unknown chlorohydrocarbon has $3.55\%$ of chlorine. If each molecule of the hydrocarbon has one chlorine atom only,the number of chlorine atoms present in $1\,g$ of chlorohydrocarbon is (Atomic wt. of $Cl = 35.5\,u$; Avogadro constant $= 6.023 \times 10^{23}\,mol^{-1}$)
- A$6.023 \times 10^9$
- B$6.023 \times 10^{23}$
- C$6.023 \times 10^{21}$
- D$6.023 \times 10^{20}$
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Similar Questions
$A$ sample of $NaNO_3$ weighing $0.38 \ g$ is placed in a $50.0 \ mL$ volumetric flask. The flask is filled to the mark with water. What is the molarity of the solution?
Medium
View SolutionMatch the following and select the correct option:
List-$I$ List-$II$ (At $STP$) $(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$ $(i)$ $0.224 \ L \ CO_2$ $(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$ $(ii)$ $4.48 \ L \ CO_2$ $(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$ $(iii)$ $0.448 \ L \ CO_2$ $(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$ $(iv)$ $2.24 \ L \ CO_2$ $(v)$ $22.4 \ L \ CO_2$
| List-$I$ | List-$II$ (At $STP$) |
| $(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$ | $(i)$ $0.224 \ L \ CO_2$ |
| $(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl}$ | $(ii)$ $4.48 \ L \ CO_2$ |
| $(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$ | $(iii)$ $0.448 \ L \ CO_2$ |
| $(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$ | $(iv)$ $2.24 \ L \ CO_2$ |
| $(v)$ $22.4 \ L \ CO_2$ |
How many $mL$ of $1 \, M$ $H_2SO_4$ is required to neutralize $10 \, mL$ of $1 \, M$ $NaOH$ solution?
Medium
View SolutionVolume of water needed to mix with $10 \ mL$ of $10 \ N \ HNO_3$ to get $0.1 \ N \ HNO_3$ is ............ $mL$.
DifficultAIIMS 2017
View SolutionWhat is the normality of $1.5 \ M \ H_3PO_4$ (in $.5$)?
Easy
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