The number of ${C O}_{2}$ molecule in $100 {c m}^{3}$ of gas at STP are:

2.688 \times 10^{21}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6.022 \times 10^{21}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.022 \times 10^{25}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.23 \times 10^{20}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $2.688 \times 10^{21}$
No.of moles $= \frac{V o l u m e}{22.4}$
$\frac{100}{22.4} \times \frac{1}{1000}$
$= \frac{1}{224}$
$≃ 0.0045 m o l e s$
$≃ 4.5 \times 10^{- 3} \times 6.023 \times 10^{23}$
$≃ 2.7 \times 10^{21} t i m e s$

Suggest Corrections

Similar questions

Q. The number of molecules in 89.6 L of a gas at STP are

Find the number of moles in 6.022 ×10²⁵molecules of CO₂

Q. If

1 mL

of water contains 20 drops. Then no. of molecules in a drop of water is
(Given

N_{A} = 6.022 \times 10^{23}

)

Question 9
3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) $6.68 \times 10^{23}$
(b) $6.09 \times 10^{22}$
(c) $6.022 \times 10^{23}$
(d) $6.022 \times 10^{21}$

Q. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a)

6.68 \times 10^{23}

(b)

6.09 \times 10^{22}

(c)

6.022 \times 10^{23}

(d)

6.022 \times 10^{21}

Join BYJU'S Learning Program

The number of CO2 molecule in 100cm3 of gas at STP are:

The correct option is A 2.688×1021No.of moles =Volume22.410022.4×11000=1224≃0.0045 moles≃4.5×10−3×6.023×1023≃2.7×1021 times

The number of ${C O}_{2}$ molecule in $100 {c m}^{3}$ of gas at STP are:

The correct option is A $2.688 \times 10^{21}$
No.of moles $= \frac{V o l u m e}{22.4}$
$\frac{100}{22.4} \times \frac{1}{1000}$
$= \frac{1}{224}$
$≃ 0.0045 m o l e s$
$≃ 4.5 \times 10^{- 3} \times 6.023 \times 10^{23}$
$≃ 2.7 \times 10^{21} t i m e s$