Question

Class - 9

Chemistry...


Calculate the number of oxygen atom in 30.3 g of Potassium nitrate ?


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Answer 1

Hi there !

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Solution :

Potassium nitrate is written as KNO3.

Molar mass of KNO3 is

K -- 1 × 39 = 39

N -- 1 × 14 = 14

O -- 3 × 16 = 48

.·. KNO3 = ( 39 + 14 + 48)

=> 101 g.

Now,

101g of KNO3 = 1 mole

1 g of KNO3 = 1 / 101 moles.

30.3 g of KNO3 = 1/101 × 30.3

=> 0.3 moles.

Now,

1 mole of KNO3 = 3 mole of oxygen atom.

0.3 mole of KNO3 = 0.3 × 3

=> 0.9 mole.

Number of oxygen atom

=> 0.9 × 6.023 × 10²³

=> 5.4207 × 10²³

Hence,

No. of oxygen atom = 5.4207 × 10²³.

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Thanks for the question !

Answer 2

Here is your answer

$\huge\underline{\red{\bold{Kn03-pottasium nitrate}}}$

$\boxed{\green{\bold{we\:find \:molar\: mass \:of\: element }}}$

The molar mass of element In kno3

k - 1×39= 39
N - 1×14=14
O3 - 3×16=48

Total weight in kno3
=> [ 39 + 14 + 48=101g ]

$\huge\boxed{\orange{\bold{Now}}}$

101 gm of kno3 = 1 mole

$1 \: gram \: of \: kno3 \: = \frac{1}{101} mole$
$30.3 \: gram \: of \: kno3 \: = \frac{1}{101} \times 30.3mole \: \\ = 0.3mole$

$\huge\boxed{\pink{\bold{Now}}}$

one mole of kno3 = 3 moles of oxygen atom

0.3 mole of know = 0.3× 3 = 0.9 mole

Number of oxygen atom

$= 0.9 \times 6.023 \times 10 {}^{23} \\ = 5.4207 \times 10 {}^{23}$
so,
$number \: of \: oxygen \: atom \\ \huge\boxed{\red{\bold{5.4207 \times10 {}^{23}}}}$


Hope it helps you