Atomic weight of an element is $x$. The actual mass of one atom of that element is
- A$x \ g$
- B$x \ amu$
- C$x \times 6.023 \times 10^{23} \ amu$
- D$\frac{x}{6.023 \times 10^{23}} \ amu$
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Similar Questions
$0.45 \ g$ of an acid (mol wt. $= 90$) required $20 \ mL$ of $0.5 \ N$ $KOH$ for complete neutralization. The basicity of the acid is:
Medium
View SolutionWhich phrase would be incorrect to use?
Easy
View SolutionCalculate the average atomic mass of chlorine using the following data:
Isotope Natural abundance (%) Molar mass $(u)$ $^{35}Cl$ $75.77$ $34.9689$ $^{37}Cl$ $24.23$ $36.9659$
(in $u$)
| Isotope | Natural abundance (%) | Molar mass $(u)$ |
|---|---|---|
| $^{35}Cl$ | $75.77$ | $34.9689$ |
| $^{37}Cl$ | $24.23$ | $36.9659$ |
Medium
View SolutionAmong $10^{-9} \ g$ (each) of the following elements,which one will have the highest number of atoms?
Element: $Pb, Po, Pr$ and $Pt$
Element: $Pb, Po, Pr$ and $Pt$
Dulong and Petit's law is valid only for
Medium
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