Calculate the number of molecules in 3.5 gms of carbon monoxide at 273 K and 760 mm pressure.

[Avogadro's number $=$ 6 x 10 $^{23}$ ]

7.5

10^{23}

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7.5

10^{22}

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1.5

10^{24}

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1.5

10^{23}

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Solution

The correct option is B $7.5$ x $10^{22}$
$28$ grams of $C O$ contain $6.023 \times 10^{23}$ molecules

$3.5$ grams of $C O$ contans $x$ molecules.

$x = \frac{3.5 \times 6.023 \times 10^{23}}{28} = 0.752 \times 10^{23}$

$= 7.52 \times 10^{22}$ .

Hence, option B is correct.

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Calculate the number of molecules in 3.5 gms of carbon monoxide at 273 K and 760 mm pressure. [Avogadro's number = 6 x 1023 ]

The correct option is B 7.5 x 102228 grams of CO contain 6.023×1023 molecules3.5 grams of CO contans x molecules. x=3.5×6.023×102328=0.752×1023 =7.52×1022.Hence, option B is correct.

Calculate the number of molecules in 3.5 gms of carbon monoxide at 273 K and 760 mm pressure.

[Avogadro's number $=$ 6 x 10 $^{23}$ ]

The correct option is B $7.5$ x $10^{22}$
$28$ grams of $C O$ contain $6.023 \times 10^{23}$ molecules

$3.5$ grams of $C O$ contans $x$ molecules.

$x = \frac{3.5 \times 6.023 \times 10^{23}}{28} = 0.752 \times 10^{23}$

$= 7.52 \times 10^{22}$ .

Hence, option B is correct.