If 24 times 10^{23} molecules are removed fro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The molar mass of hydrogen gas $(H_2)$ is $2 \ g/mol$.The initial number of moles is calculated as: $n = \frac{50 \ g}{2 \ g/mol} = 25 \ mol$.The

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(A) The molar mass of hydrogen gas $(H_2)$ is $2 \ g/mol$.The initial number of moles is calculated as: $n = \frac{50 \ g}{2 \ g/mol} = 25 \ mol$.The number of moles removed is calculated using Avogadro's number $(N_A = 6 	imes 10^{23} \ mol^{-1})$: $n_{removed} = \frac{24 	imes 10^{23}}{6 	imes 10^{23}} = 4 \ mol$.The remaining moles of hydrogen are: $25 \ mol - 4 \ mol = 21 \ mol$.

If $24 \times 10^{23}$ molecules are removed from $50 \ g$ of hydrogen gas $(H_2)$,how many moles of hydrogen remain?

Similar Questions

Which one of the following will have the largest number of atoms?
$(i)$ $1 \,g$ $Au_{(s)}$
$(ii)$ $1 \,g$ $Na_{(s)}$
$(iii)$ $1 \,g$ $Li_{(s)}$
$(iv)$ $1 \,g$ of $Cl_{2(g)}$

Which of the following pairs of gases contains the same number of molecules?

What is the number of atoms present in $16 \ g$ of oxygen gas $(O_2)$?

$4.4 \ g$ of $CO_2$ and $2.24 \ L$ of $H_2$ at $\text{STP}$ are mixed in a container. The total number of molecules present in the container will be :

How will you differentiate $0.50 \ mol$ $Na_2CO_3$ and $0.50 \ M$ $Na_2CO_3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module