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Standard XII

Physics

Law of Radioactivity

If 32 g of O ...

Question

If 32 g of $O_{2}$ contains 6.022 $\times 10^{23}$ molecules at NTP then 32 g of S, under the same conditions, will contain,

6.022 x 10²³ S

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3.011 x 10²³ S

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12.044 x 10²³ S

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1 x 10²³ S

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Solution

The correct option is A
6.022 x 10²³ S

$n_{s}$ = $\frac{32}{32}$ =1

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Similar questions

Q. If

32

g of

O_{2}

contains

6.022 \times 10^{23}

molecules at NTP then,

32

g of

S

under the same conditions will contain :

One mole of any substance contains $6.022 \times 10^{23}$ atoms/molecules. Number of molecules of $H_{2} S O_{4}$ present in 100 mL of $0.02 M H_{2} S O_{4}$ solution is.............

(a) $12.044 \times 10^{20}$ molecules (b) $6.022 \times 10^{23}$ molecules

One mole of oxygen gas at STP is equal to...........

(a) $6.022 \times 10^{23}$ molecules of oxygen

(b) $6.022 \times 10^{23}$ atoms of oxygen

(d) 32 g of oxygen

Q. 12 g C-12 contains

6.022 \times 10^{23}

atoms of carbon.
(a)

6.022 \times 10^{23}

is known as .............
(b) Calculate the number of carbon atoms present in 48 g C-12.
(c) Which weighs more,

6.022 \times 10^{23}

molecules of

C O_{2}

6.022 \times 10^{23}

molecules of

H_{2} O

How many molecules are present in 9g of water?

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If 32 g of O2 contains 6.022 ×1023 molecules at NTP then 32 g of S, under the same conditions, will contain,

The correct option is A 6.022 x 1023 S ns =3232 =1

If 32 g of $O_{2}$ contains 6.022 $\times 10^{23}$ molecules at NTP then 32 g of S, under the same conditions, will contain,

The correct option is A
6.022 x 10²³ S

$n_{s}$ = $\frac{32}{32}$ =1