Question

If $2.77\times {10}^{25}$ electrons are there in a $MgC{l}_2$ sample, then what is the mass of $MgC{l}_2$ present?

• A. $95\text{g}$
• B. $190\text{g}$
• C. $45\text{g}$
• D. $110\text{g}$

Answer 1

The correct option is A $95\text{g}$

The number of electrons in one molecule of $MgC{l}_2=(12+(17\times 2\left)\right)=46$
So, the number of electrons in one mol of $MgC{l}_2=46\times 6.022\times {10}^{23}$
Therefore,
$46\times 6.022\times {10}^{23}$ electrons are contained in 1 mol of $MgC{l}_2$
Hence $2.77\times {10}^{25}$ electrons will be contained in = $\frac{2.77\times {10}^{25}}{46\times 6.022\times {10}^{23}}$ mol of $MgC{l}_2$
= 0.999 mol of $MgC{l}_2$
Thus, the required mass of $MgC{l}_2=(24+2\times 35.5)\times 0.999\text{g}$
$=94.905\text{g}\cong 95\text{g}$