Which of the following contains the maximum n | vedclass

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

Question

(C) The number of molecules is proportional to the number of moles $(n)$.$(A)$ $n = \frac{0.4 	ext{ g}}{2 	ext{ g/mol}} = 0.2 	ext{ mol}$.$(B)$ $n

Vedclass · Accepted Answer

$8 	ext{ dm}^3 	ext{ of } H_2 	ext{ gas at STP}$

Vedclass · Answer

(C) The number of molecules is proportional to the number of moles $(n)$.$(A)$ $n = \frac{0.4 	ext{ g}}{2 	ext{ g/mol}} = 0.2 	ext{ mol}$.$(B)$ $n = \frac{2 	ext{ g}}{32 	ext{ g/mol}} = 0.0625 	ext{ mol}$.$(C)$ At $STP$, $1 	ext{ mol}$ occupies $22.4 	ext{ dm}^3$. So, $n = \frac{8 	ext{ dm}^3}{22.4 	ext{ dm}^3/	ext{mol}} \approx 0.357 	ext{ mol}$.$(D)$ $n = \frac{2 	ext{ dm}^3}{22.4 	ext{ dm}^3/	ext{mol}} \approx 0.089 	ext{ mol}$.Comparing the values, $0.357 	ext{ mol}$ is the highest. Thus, option $(C)$ contains the maximum number of molecules.

Which of the following contains the maximum number of molecules?

Similar Questions

$A$ sample of ammonium phosphate,$(NH_4)_3PO_4$,contains $6$ moles of hydrogen atoms. The number of moles of oxygen atoms in the sample is:

Arrange the following in order of increasing mass:
$I$. $1$ mole of $N_2$
$II$. $0.5$ mole of $O_3$
$III$. $3.011 \times 10^{23}$ molecules of $O_2$
$IV$. $0.5$ gram atom of $O$

Calculate the number of valence electrons in $4.2 \ g$ of $N_3^-$ ions in terms of $N_A$ ($N_A$ = Avogadro constant). (in $N_A$)

$7.0 \, g$ of $N_2$,$8.0 \, g$ of $O_2$,and $34.0 \, g$ of $NH_3$ are present in a $1 \, L$ vessel. The active masses in $mol \, L^{-1}$ are respectively:

Find the maximum number of atoms in the following:
$(i)$ $1 \ g$ $Au_{(s)}$
$(ii)$ $1 \ g$ $Na_{(s)}$
$(iii)$ $1 \ g$ $Li_{(s)}$
$(iv)$ $1 \ g$ $Cl_{2(g)}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module