Question

how many molecules of water are present in 75 g of H2O

Answer 1

Answer:

There are 2.342558 X 10^24 molecules of water in a 75.0-gram sample of H2 O.

Explanation:

We have,

1amu =1÷(Avogadro’s number)= 1÷(6.023×10^23) = 1.66 × 10^(-24) g

(You may or may not know)

1 amu = 1.66 × 10^(-24) grams —————— (i)

Given, mass of H20 = 75 g

We know, Molecular mass of H2O=2×1+(16×1)amu= 18 amu

As, 1 amu= 1.66×10^(-24) g

So, (1 ÷ ( 1.66×10^(-24) ) ) amu = 1g

Then, 75 g = 75 × (1 ÷ ( 1.66×10^(-24) ) ) amu = 4.51725 × 10^25 amu.

Since, 1 molecule of H2O weighs = 18 amu.

1 amu = (1 ÷ 18 ) molecule of H2O

4.51725 × 10^25 amu = (1 ÷ 18 ) × 4.51725 × 10^25 molecules of H20

= 2.50958333 × 10^24 molecules of H20.

Hence, 75 g of H20 contains 2.50958333 × 10^24 molecules of H20.

This process might be a bit confusing but once you grasp the concept its easy.

Answer 2

The number of molecules of water are present in 75 g of $H_{2} O$ is $25.1159 \times 10^{23}$.

Explanation:

Given:

75 g of $H_{2} O$.

To Find:

The number of molecules of water are present in 75 g of $H_{2} O$.

Solution:

As given,75 g of $H_{2} O$.

The formula of water is $H_{2} O$.

The atomic mass of $H_{2} =2\times1=2$ and the atomic mass of $O=1\times16=16$.

The molecular mass of $H_{2} O=2+16=18$.

The number of moles in 75 g of water $=\frac{75}{18} = 4.17 moles$

According to Avogadro's number 1 mole$=6.023\times 10^{23}$ molecules

Therefore,the number of molecules in  4.17 mole

$=4.17\times 6.023\times 10^{23} =25.1159 \times 10^{23}$ molecules

Thus,the number of molecules of water are present in 75 g of $H_{2} O$ is $25.1159 \times 10^{23}$.

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