Question

( 70 cdot 5 ) moles of ( H_{2} 0=frac{3.6}{18}=0.2 ) moles.
How; Imole ( y ) electron ( =6.02 times 10^{23} ) 0.2 mole ( frac{5}{7} e^{-1}=0.2 times 6.02 times 10^{23} )
( 1.204 times 10^{23} )

# 25. Number of electrons in 3.6 grams of H2O are (1) 12.04 × 1023 (2) 6.02 × 1023 (3) 1.204 x 1023 (4) 120.44 x 10-23

Solution