The number of water molecules present in a drop of water (volume 0.0018 mL) density = 18 mL-1 at room temperature is

Question

The number of water molecules present in a drop of water (volume 0.0018 mL) density = 18 mL-1 at room temperature is (A) 1.084 × 1018  (B) 6.023 × 1019 (C) 4.84 × 1017  (D) 6.023 × 1023

Tardigrade Admin · Accepted Answer

0.0018 mL = 0.0018 g = 0.0001 mole of water = 1 × 10-4 mole. ∴ number of water molecules = 6.023 × 1023 × 10-4 = 6.023 × 1019

Q. The number of water molecules present in a drop of water (volume 0.0018 mL) density = 18 $m L^{- 1}$ at room temperature is

$1.084 \times 1 0^{18}$

$6.023 \times 1 0^{19}$

$4.84 \times 1 0^{17}$

$6.023 \times 1 0^{23}$

$0.0018 m L = 0.0018 g = 0.0001$ mole of water
$= 1 \times 1 0^{- 4} m o l e$ .
$∴$ number of water molecules
$= 6.023 \times 1 0^{23} \times 1 0^{- 4}$
$= 6.023 \times 1 0^{19}$

Q. The number of water molecules present in a drop of water (volume 0.0018 mL) density = 18 mL−1 at room temperature is

1.084×1018

6.023×1019

4.84×1017

6.023×1023