Question

How many H-atoms are present in $0.046g$ of ethanol?

• A. $6\times 10^{20}$
• B. $1.2\times 10^{21}$
• C. $3\times 10^{21}$
• D. $3.6\times 10^{21}$

Answer 1

Answer: (D)

Mol. wt. of $C_2{H}_{5}OH$
$=2\times 12+5+16+1=64$
$\because 48g{C}_2{H}_{5}OH$ has H atom $=6\times N_A$
$\therefore 0.046g{C}_2{H}_{5}OH$ has H atom
$=\frac{6\times 6.02\times 10^{23}\times 0.046}{46}$
$=3.6\times 10^{21}$