Question

The number of protons present in 90ml of water is

Answer 1

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Answer 2

Answer : The number of protons present in 90 ml of water is $30.055\times 10^{24}$ protons.

Solution : Given,

Volume of water = 90 ml

Density of water = 0.9982 g/ml

Molar mass of water = 18 g/mole

First we have to calculate the mass of water.

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=(0.9982g/ml)\times 90ml=89.838g$

Now we have to calculate the moles of water.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{89.838g}{18g/mole}=4.991moles$

Now we have to calculate the molecules of water.

1 mole of water has $6.022\times 10^{23}$ molecule of water

4.991 moles of water has $4.991\times 6.022\times 10^{23}=30.055\times 10^{23}$ molecule of water

Now we have to calculate the number of protons in water.

The number of protons present in 1 molecule of water is,

$H_2O$ = 2(1) + 8 = 10 protons

Number of protons present in 1 molecule of water= 10 protons

Number of protons present in $30.055\times 10^{23}$ molecule of phosphorous = $10\times 30.055\times 10^{23}=30.055\times 10^{24}$ protons

Therefore, the number of protons present in 90 ml of water is $30.055\times 10^{24}$ protons.