Question

From $2mg$ calcium $1.2\times {10}^{19}$ atoms are removed. The amount of calcium left is:
(Take $N_A=6\times {10}^{23}$)

• A. $2\times {10}^{19}$ atoms
• B. $2\times {10}^{-5}$ mol
• C. $3\times {10}^{-5}$ mol
• D. $3\times {10}^{19}$ atoms

Answer 1

The correct option is C $3\times {10}^{-5}$ mol

Moles of $Ca=\frac{2\times {10}^{-3}}{40}=0.05\times {10}^{-3}$
no. of atoms of $Ca=0.05\times {10}^{-3}\times 6\times {10}^{23}$
$=0.3\times {10}^{20}$
$=3\times {10}^{19}$
no. of atoms removed = $1.2\times {10}^{19}$
no. of atoms left = $3\times {10}^{19}-1.2\times {10}^{19}$
$=[3-1.2]\times {10}^{19}$
$=1.8\times {10}^{19}$
So moles left$=\frac{1.8\times {10}^{19}}{6\times {10}^{23}}=0.300\times {10}^{-4}$
$=3.0\times {10}^{-5}mol$