Question
Calcium Carbonate
✓
Answer
Given: Number of moles of Calcium carbonate $\left( CaCO _3\right) n =0.2 mol$
To find: Mass in grams of $0.2$ mol of $CaCO _3$
Solution:
Molecular mass of $\left( CaCO _3\right) M$
$=($ Atomic mass of Ca $) \times 1+($ Atomic mass of C $) \times 1+($ Atomic mass of O $) \times 3$
$=(40 \times 1)+(12 \times 1)+(16 \times 3)$
$=40+12+48$
Molecular mass of $\left( CaCO _3\right) M =100$
According to the formula Number of moles in the given $CaCO _3( n )$
$\therefore$ Mass of $CaCO _3$ in grams $( m )=0.2 \times 100$
$\therefore$ Mass of $CaCO _3$ in grams $( m )=20 g$
Mass of $0.2$ mole of $CaCO _3$ is $20\ g$
To find: Mass in grams of $0.2$ mol of $CaCO _3$
Solution:
Molecular mass of $\left( CaCO _3\right) M$
$=($ Atomic mass of Ca $) \times 1+($ Atomic mass of C $) \times 1+($ Atomic mass of O $) \times 3$
$=(40 \times 1)+(12 \times 1)+(16 \times 3)$
$=40+12+48$
Molecular mass of $\left( CaCO _3\right) M =100$
According to the formula Number of moles in the given $CaCO _3( n )$
$\therefore$ Mass of $CaCO _3$ in grams $( m )=0.2 \times 100$
$\therefore$ Mass of $CaCO _3$ in grams $( m )=20 g$
Mass of $0.2$ mole of $CaCO _3$ is $20\ g$
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