Question

A drop of water has a volume of approximately 7×10−2 ml. How many water molecules does it contain? The density of water is 1.0 g/cm3.

Answer 1

The volume of drop of water is $7\times 10^{-2} mL$ and density of water is $1 g/cm^{3}$.

Since, $1 mL=1 cm^{3}$

Thus, volume will be  $7\times 10^{-2} cm^{3}$.

Mass of drop of water can be calculated from density and volume as follows:

$m=d\times V=1 g/cm^{3}\times 7\times 10^{-2}cm^{3}=7\times 10^{-2}g$

To calculate number of water molecules first calculate the number of moles as follows:

$n=\frac{m}{M}$

Molar mass of water is 18 g/mol thus,

$n=\frac{7\times 10^{-2}g}{18 g/mol}=0.00388 mol$

Now, in 1 mol of a substance there are $6.023\times 10^{23}$ molecules thus, number of molecules in 0.00388 mol will be:

$N=0.00388 mol\times 6.023\times 10^{23}=2.342\times 10^{21} molecules$

Therefore, number of water molecules in drop will be $2.342\times 10^{21}$.

Answer 2

The number of molecules present in 7×10¯² mL of water is 2.34×10²¹ molecules  

• We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 7×10¯² mL

Density of water = 1 g/mL

Mass of water =?

Mass = Density × Volume

Mass of water = 1 × 7×10¯²

Mass of water = 0.07 g

• Finally, we shall determine the number of molecules present in 0.07 g of water

From Avogadro's hypothesis,

1 mole of H₂O = 6.02×10²³ molecules

But

1 mole of water (H₂O) = (1×2) + 16 = 18 g

Thus,

18 g of H₂O = 6.02×10²³ molecules

Therefore,

0.07 g of H₂O = $\frac{0.07 * 6.02*10^{23}}{18}$ = 2.34×10²¹ molecules

Thus, 2.34×10²¹ molecules are present in 7×10¯² mL of water.

Learn more: https://brainly.com/question/9542350