Question

How many grams of nitrogen in 7.3 x 1024 molecules of NaNO3 ?
Group of answer choices

1.4 x 102 g N

.014 g N

1.7 x 102 g N

0.017 g N

Answer 1

Answer:

Given

Number of molecules of NaNO3 = 7.3 * 10²⁴

To find

Weight of Nitrogen in corresponding ammount of NaNO3 = ?

Solution

$Number \: of \: moles \: of \: NaNO3 = \frac{No \: of \: molecules}{Avagadro's \: number}

$

$Number \: of \: moles \: of \: NaNO3 = \frac{7.3 \: * \: 10²⁴}{6.023 \: * \: 10²³} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 12.12 \: moles

$

Now,

Number of moles of NaNO3 = Number of moles of Nitrogen,

Number of moles of Nitrogen atom = 12.12 moles

Mass of 1 mole nitrogen atom = 7 g

Mass of 12.12 moles nitrogen atom = 12.12×7 = 84.84 grams

$\fbox{84.84 grams}$

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