# Which option contains the same number of chlorine atoms as present in a 142*g mass of Cl_2?

## $\text{a. 222 g calcium chloride;}$ $\text{b. 111 g magnesium chloride;}$ $\text{c. 444 g calcium chloride;}$ $\text{d. 234 g sodium chloride?}$

$\text{Moles of}$ $C {l}_{2} = \frac{142 \cdot g}{70.9 \cdot g \cdot m o {l}^{-} 1} = 2.00 \cdot m o l$
Now $N a C l$ has a molar mass of $58.44 \cdot g \cdot m o {l}^{-} 1$
And thus we need a FOUR MOLE quantity of this salt, i.e. $2 \cdot m o l \times 58.44 \cdot g \cdot m o {l}^{-} 1 = 234 \cdot g$.