A quantity of aluminium has a mass of 54.0 g | vedclass

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

Question

(C) Step $1$: Calculate the number of moles of $Al$ atoms.$n(Al) = \frac{	ext{mass of } Al}{	ext{molar mass of } Al} = \frac{54.0 \ g}{27.0 \ g/mol}

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(C) Step $1$: Calculate the number of moles of $Al$ atoms.$n(Al) = \frac{	ext{mass of } Al}{	ext{molar mass of } Al} = \frac{54.0 \ g}{27.0 \ g/mol} = 2.0 \ mol$.Step $2$: Since the number of atoms is the same,the number of moles of $Mg$ atoms must also be $2.0 \ mol$.Step $3$: Calculate the mass of $Mg$ atoms.$	ext{mass of } Mg = n(Mg) 	imes 	ext{molar mass of } Mg = 2.0 \ mol 	imes 24.0 \ g/mol = 48.0 \ g$.

$A$ quantity of aluminium has a mass of $54.0 \ g$. What is the mass of the same number of magnesium atoms? ........... $g$

Similar Questions

Calculate the number of molecules present in $5.4 \ g$ of urea (Molar mass of urea $= 60 \ g \ mol^{-1}$).

If we take $44 \ g$ of $CO_2$ and $14 \ g$ of $N_2$,what will be the mole fraction of $CO_2$ in the mixture?

Calculate the mass of oxygen present in $0.1 \ mol$ of $Na_2CO_3 \cdot 10H_2O$. (in $g$)

The number of atoms in $8 \ g$ of sodium is $x \times 10^{23}$. The value of $x$ is ...... (Nearest integer).
$[ \text{Given}: N_{A} = 6.02 \times 10^{23} \ mol^{-1}, \text{Atomic mass of } Na = 23.0 \ u ]$

$8 \ g$ of $O_2$ has the same number of atoms as that in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module