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How many grams of Ba(NO3)2 contain 5.4 x 10^24

User Mhmt
by
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2 Answers

2 votes

Answer:

2352.15g or 2.4kg (significant digits)

Step-by-step explanation:

Assuming that it's 5.4*10^24 molecules of Ba(NO3)2, We know that a mole has 6.02* 10^23 molecules. Dividing the given number by Avagadro's number, we can determine how many mols are there.

(5.4*10^24)/(6.02*10^23)=8.970099668...

lets round this to 9.0 moles (Significant Digits.)

now knowing this, we can take the atomic mass of the compound(I measure using my school's given masses, may be different for you.)

Ba=137.33g/mol

N=14.01g/mol

O=16.00g/mol

Add these up

137.33+(28.02+96.00)=261.35g/mol

now multiply the atomic mass by moles.

261.35*9=2352.15g or 2.4kg(significant digits)

I hope this helps

User Vinnie Fusca
by
3.5k points
4 votes

Step-by-step explanation:

Solution given:

1 mole of Ba(NO3)2=261 g

1 mole of Ba(NO3)2=6.023×
{10}^(3)molecules

we have:

6.023×
{10}^(23) molecules=261g

5.4×
{10}^(24) molecules =
\frac{261}{6.023 * {10}^(23) } * 5.4*{10}^(24) = 2340gis your answer.

User Esa Hannila
by
3.6k points