Question

A drop of water has a volume of approximately 7x10-2 mL

Answer 1

Solution:-

Volume (V) of water drop = 7 × 10^-2 ml

Density (d) of water = 1 g/cm³ = 1 g/mL ( $\because$ 1 cm³ = 1 mL)

​​​​​​Thus, mass (m) of water drop = Density × Volume = (1 g/mL) × (7 × 10^-2 ml) = 7 × 10^-2 g = 0.07 g

Now,

N = (m / M) × N​​​​​​​₀

_{​​​​​}Where,

N = Number of water molecules

m = mass of water drop

M = molar mass of water (= 18 g/mole)

N₀ = Avogadro's Number (= 6.022 × 10²³)

N = (0.07 / 18) × (6.022 × 10²³)

N = 2.34 × 10²¹

Hence, drop contains = 2.34 × 10²¹ molecules of water.

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