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3 years ago

What is the number of moles of Sodium Chloride (NaCl) that is present in 29.22g sample of it?

Chemistry

2 answers:

kodGreya [7K]3 years ago

5 0

Answer:

$\boxed {\boxed {\sf 0.5000 \ mol \ NaCl}}$

Explanation:

We are asked to convert grams to moles. We must use the molar mass or the grams in 1 mole of a substance. These values are found on the Periodic Table as the atomic mass, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound sodium chloride or NaCl. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
Cl: 35.45 g/mol

The formula for the compound does not contain any subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the molar masses together to find sodium chloride's molar mass.

NaCl: 22.9897693 + 35.45= 58.4397693 g/mol

Now we can use the molar mass to create a ratio. We know there are 58.4397693 grams of NaCl in 1 mole of NaCl.

$\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

Multiply by the number of grams provided in the problem: 29.22

$29.22 \ g\ NaCl*\frac {58.4397693 \ g \ NaCl}{ 1 \ mol \ NaCl}$

Flip the ratio so the units of grams of NaCl cancel.

$29.22 \ g\ NaCl*\frac { 1 \ mol \ NaCl}{58.4397693 \ g \ NaCl}$

$29.22*\frac { 1 \ mol \ NaCl}{58.4397693}$

$\frac {29.22}{58.4397693} \ mol \ NaCl$

$0.5000019738 \ mol \ NaCl$

The original measurement of grams has 4 significant figures, so our answer must have the same. For our answer, that is the ten-thousandth place. The 0 to the right of this place (0.5000019738) tells us to leave the 0 in the ten-thousandth place.

$0.5000 \ mol \ NaCl$

There are approximately <u>0.5000 moles of sodium chloride in a 29.22 gram sample of sodium chloride.</u>

Liula [17]3 years ago

3 0

There are 0.5 moles of NaCl present in a 29.22g sample.

to find moles of a compound you would look at the periodic table to find the mass. in this case, the mass of NaCl is 58.44 g/mol. now, you divide the sample mass of 29.22g by 58.44 g/mol. the grams will cancel each other out and you’re left with 0.5 mol in that sample

hope this helps!

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