A) 4.0 g of hydrogen
B) 71.0g chlorine
C) 127.0 g of iodine
D) 48.0g magnesium
Correct Answer: A
Solution :
4g of hydrogen =4 mole of hydrogen \[=4\times 6.023\times {{10}^{23}}\]atoms 71.0 gm of chlorine \[=\frac{71.0}{71.0}=1\]moles of chlorine\[=6.023\times {{10}^{23}}\] \[=\frac{127}{254}=6.023\times {{10}^{23}}\times \frac{1}{2}\]48.0 gm of magnesium \[=\frac{48.0}{24.0}=2\times 6.023\times {{10}^{23}}\] \[\therefore \]4.0 gm \[{{H}_{2}}\] has largest number of atoms.
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