How many electrons are present in 3.6 g of water (H2O, M = 18)? Take N_A = 6.02x10^23 mol^-1. Give the answer in units of 10^23, to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3mc126

How many electrons are present in 3.6 g of water ( $H X_{2} O$ , $M = 18$ )? Take $N_{A} = 6.02 \times 1 0^{23} mo l^{- 1}$ . Give the answer in units of $1 0^{23}$ , to two decimal places.

Numerical answer type, answer in x10^23, tolerance 0.02.

Moles of water

= 3.6/18 = 0.20

. Each

H X_{2} O

molecule has

2 (1) + 8 = 10

electrons. Electrons

= 0.20 \times 10 \times 6.02 \times 1 0^{23} = 12.04 \times 1 0^{23}