Question

17.
The number of water molecules present in a drop of
water (volume = 0.0018 ml) at room temperature is
(density of H,0 = 1 g/mL)
(a) 6.023 x 1019 (b) 1.084 1018
(c) 4.84 x 1017
(d) 6.023 x 1023​

Answer 1

At room temperature I.e. 25°c,the density of water is 1g/L.
Mass of water=density × volume
=1g/L ×0.0018ml
=1.8×10^-3 g.
18g of water =1 mole
1 mole of water contains 6.023 ×10^23 molecules of water.
18g of water contains 6.023×10^23 molecules of water.
So, 1.8 ×10^-3 g of water contains
=1.8×10^3×6.023 10^23 /1.8
=6.023×10^19.