Question # Number of electrons in ( 1.8 mathrm{mL} ) of ( mathrm{H}_{2} mathrm{O}(ell) ) is about:

# Number of electrons in ( 1.8 mathrm{mL} ) of ( mathrm{H}_{2} mathrm{O}(ell) ) is about:

(A) ( 6.02 times 10^{23} )

(B) ( 3.011 times 10^{23} )

(C) ( 0.6022 times 10^{21} )

(D) ( 60.22 times 10^{20} )

Solution

( 1.8 mathrm{me} ) od ( mathrm{H}_{2} mathrm{O} )

no of electrom

[

frac{118}{180}=frac{1}{10}

]

electrom = mole x ( N_{a} )

[

begin{array}{l}

frac{1}{10} times 6.02 times 10^{23}=

=6.02 times 10^{22}

end{array}

]