You visited us 1 times! Enjoying our articles? Unlock Full Access!

Molecular Mass

Substance hav...

Question

Substance having maximum number of molecules is:

44.8 L

H_{2}

at S.T.P.

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

2

g of

H e

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

16

g of

C H_{4}

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

33

g of

C O_{2}

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

Open in App

Solution

The correct option is A $44.8 L$ of $H_{2}$ at S.T.P.
1 mole of a gas at STP $= 22.4 L$

$44.8$ L of hydrogen at STP corresponds to $\frac{44.8}{22.4} = 2$ moles.

$2$ g helium corresponds to $\frac{2}{4} = 0.5$ moles.

$16$ g methane corresponds to $\frac{16}{16} = 1$ mole.

$33$ g carbon dioxide corresponds to $\frac{33}{44} = 0.75$ moles.

Hence, option A is the correct answer.

Suggest Corrections

Similar questions

The volumes of gases $H_{2}, C H_{4}, C O_{2} a n d N H_{3}$ absorbed by 1 g of charcoal at 288 K are in the order:

C O_{2}

16

C H_{4}

48

O_{2}

C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O

Q. 1 mole of CH4 on combustion gave 33g CO2 thus % yield is

3.0115 \times 10^{23}

(P_{4})

10

H_{2}

1 g

N a_{2} C O_{3}

33.6 L

C O_{2}

C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O

16 g

C H_{4}

22.4 L

O_{2}

Join BYJU'S Learning Program