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Standard XII

Chemistry

Mole Concept

12 g C-12 con...

Question

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 $C O_{2}$ or $6.022 \times 10^{23}$ molecules of $H_{2} O$ ?

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Solution

(a) Avogadro's number
(b) $4 \times 6.022 \times 10^{23} = 24.088 \times 10^{23}$ carbon atoms
(c) $C O_{2}$ molecules

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Similar questions

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

Q. Data about three gases at STP is given:
(A)

16 g C H_{4}

(B)

11.2 L C O_{2}

(C)

6.022 \times 10^{23}

molecules of

N H_{3}

(a) The number

6.022 \times 10^{23}

is known as .............
(b) Calculate the number of

C H_{4}

molecules present in 16 g

C H_{4}

.
(c) Arrange

A, B

and

C

in the increasing order of their mass in grams.
[given: At mass:

H = 1, C = 12, N = 14, O = 16

]

1

g of

^{12} C

contains

6.022 \times 10^{23}

atoms of Carbon.

which has more weight 6.022*1023 atoms of oxygen or carbon?

Q. Assertion :Both

138

g of

K_{2} C O_{3}

and

12

g of carbon have same number of carbon atoms. Reason: Both contain

1

g atom of carbon which contains

6.022 \times 10^{23}

carbon atoms.

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12 g C-12 contains 6.022×1023 atoms of carbon.(a) 6.022×1023 is known as .............(b) Calculate the number of carbon atoms present in 48 g C-12.(c) Which weighs more, 6.022×1023 molecules of CO2 or 6.022×1023 molecules of H2O?

(a) Avogadro's number(b) 4×6.022×1023=24.088×1023 carbon atoms(c) CO2 molecules

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 $C O_{2}$ or $6.022 \times 10^{23}$ molecules of $H_{2} O$ ?

(a) Avogadro's number
(b) $4 \times 6.022 \times 10^{23} = 24.088 \times 10^{23}$ carbon atoms
(c) $C O_{2}$ molecules