Part A) How many moles are there in a 1.40 kg bottle of water? Part B) How many molecules are...

Question:

Part A) How many moles are there in a {eq}\rm 1.40\ kg {/eq} bottle of water?

Part B) How many molecules are there in the bottle?

Mole concept:

A mole is a measure of the amount of substance contained in matter. Consider a dozen apples, a dozen bananas or a dozen anything. It all contains 12 items. Similarly, a mole of an element contains {eq}6.022 \times 10^{23} {/eq} atoms, a mole of oxygen gas contains {eq}6.022 \times 10^{23} {/eq} oxygen molecules.

One mole of an element contains {eq}W {/eq} g of mass of the element, where {eq}W {/eq} is the atomic mass of the element.

One mole of a compound contains {eq}M {/eq} g of mass of the compound, where {eq}M {/eq} is the molecular mass of the compound.

Answer and Explanation:

A.

We need to calculate the number of moles in 1.40 Kg bottle of water.

Since, 1 mole of a compound contains {eq}M {/eq} g of mass of the compound, where {eq}M {/eq} is the molecular mass of the compound.

The molecular mass of water is 18 g. One mole of water equals {eq}18 g {/eq} of water.

{eq}1 \ g \ equals \ \frac { 1 } { 18 } \ moles \ of \ water. {/eq}

1.40 Kg water equals 1400 g of water.

Thus,

{eq}\begin{align*} If \ 1 \ g \ of \ water &= \frac { 1 } { 18 } \ moles \ of \ water, \ then \\ 1400 \ g &= \frac { 1 } { 18 } \times 1400 \ moles \ of \ water \\ &= 77.78 \ moles \ of \ water. \end{align*} {/eq}

Thus, there are {eq}77.78 \ moles {/eq} in 1.4 Kg of water.

B.

One mole of a compound contains {eq}6.022 \times 10^{23} {/eq} molecules of that compound.

1.4 Kg of water equals 77.78 moles of water.

Hence the number of molecules {eq}n {/eq} in 77.78 moles of water are calculated as follows;

{eq}\begin{align*} n &= 77.78 \times 6.022 \times 10^{23} \\ &= 4.683 \times 10^{25} \ \text {molecules}. \end{align*} {/eq}

So, there are {eq}4.683 \times 10^{25} \ \text {molecules} {/eq} in the water bottle.


Learn more about this topic:

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How to Convert Grams to Moles

from Analytical Chemistry: Help & Review

Chapter 4 / Lesson 8
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