The number of water molecules present in a drop of water weighing 0.018 gm is

Question

The number of water molecules present in a drop of water weighing 0.018 gm is (A) 6.022 ×1026 (B) 6.022 ×1023 (C) 6.022 ×1019 (D) 6.022 ×1020

Tardigrade Admin · Accepted Answer

18 g of H 2 O =1 mol of H 2 O =6.022 × 1023 molecules 0.018 g of H 2 O =(0.018/18) mol of H 2 O =(0.018/18) × 6.022 × 1023 molecules =6.022 × 1020 molecules

Q. The number of water molecules present in a drop of water weighing $0.018 g m$ is

$6.022 \times 1 0^{26}$

$6.022 \times 1 0^{23}$

$6.022 \times 1 0^{19}$

$6.022 \times 1 0^{20}$

$18 g$ of $H_{2} O = 1 m o l$ of $H_{2} O = 6.022 \times 1 0^{23}$ molecules

$0.018 g$ of $H_{2} O = \frac{0.018}{18} m o l$ of $H_{2} O$

$= \frac{0.018}{18} \times 6.022 \times 1 0^{23}$ molecules

$= 6.022 \times 1 0^{20}$ molecules

Q. The number of water molecules present in a drop of water weighing 0.018gm is

6.022×1026

6.022×1023

6.022×1019

6.022×1020