# What is the mass of 9.25 times 10^22 molecules of water?

May 20, 2017

27.72g

#### Explanation:

A mole of any compound has a mass equal to the compound's relative formula mass in grams.
So, the first step is to work out how many moles of the substance there is.
There are 6.022×10^23 molecules in a mole of a substance, so to find out how many moles there are of water, divide $9.25 \times {10}^{23}$ by 6.022×10^23 which is 1.5360345... or $1.54$ to 3 significant figures.

So we have 1.54 moles of water.

Next, we need to find the relative formula mass of water using the relative atomic masses of the atoms making it up. Water is ${H}_{2} O$ so we have 2 hydrogens (atomic mass 1) and 1 oxygen (atomic mass 16). $\left(2 \times 1\right) + 16 = 18$ so water has a relative formula mass of 18.

Now we remember the rule about moles and multiply 18 by our 1.54 moles to get 27.72g (27.7 to 3SF)

Hope this helps; let me know if I can do anything else:)

May 20, 2017

$27.72$ $\text{g}$

#### Explanation:

The relationship between the number of particles and the number of mole of a substance is given by the equation $N = n L$; where $N$ is the number of particles, $n$ is the number of mole, and $L$ is Avogadro's constant.

$n$ is also defined using the equation $n = \frac{m}{M}$; where $m$ is the mass and $M$ is the molar mass of the substance.

Let's substitute the definition of $n$ into the first equation:

$R i g h t a r r o w N = n L$

$R i g h t a r r o w N = \frac{m}{M} \times L$

Then, let's substitute the value of $N$ and $L$ into the equation:

$9.25 \times {10}^{22} = \frac{m}{M} \times 6.02 \times {10}^{23}$

Now, let's evaluate the molar mass of water:

$R i g h t a r r o w M \left({H}_{2} O\right) = 2 \times 1.00 + 16.0$

$R i g h t a r r o w M \left({H}_{2} O\right) = 2.00 + 16.00$

$R i g h t a r r o w M \left({H}_{2} O\right) = 18.00$

Substituting into the equation:

$R i g h t a r r o w 9.25 \times {10}^{22} = \frac{m}{18.00} \times 6.02 \times {10}^{23}$

Dividing both sides of the equation by $6.02 \times {10}^{23}$:

$R i g h t a r r o w 1.54 = \frac{m}{18.00}$

Multiplying both sides by $18.00$:

$R i g h t a r r o w 27.72 = m$

$\therefore m = 27.72$

Therefore, the mass of $9.25 \times {10}^{22}$ molecules of water is $27.72$ $\text{g}$.