Question

Calculate the number of water molecule present in one drop of water which weigy 0.18 g.

Answer 1

Answer:

0. 18 g of water = 118 x 0.18 = 0.01 mole. 1 mole of water (Avogadro's number) contains 6.023 × 1023 water molecules.

Answer 2

Answer:6.022×10^21 molecules

Explanation:

We know that One mile of water molecules contain 6.022×10^23 water molecules .So ,in order to calculate the number of molecules in 0.18 gram of water ,we Should first calculatethe mass of one mole of water in grams . Molecular mass of water=2+16=18grams .Thus , one mole of water has a mass of 18 grams and it contains 6.022 into 10 raise to power 23 molecules of water . 18 gram of water contain=6.022×10^23 molecules; 0.18 gram water contain = (6.022 ÷18) ×0.18 = 6.022×10^21 Thus, drop of water weighing 0.18 gram contain 6.022×10^21 molecules of water in it.