Question

Determine the quantity (g) of pure CaCl2 in 7.5 g of CaCl2•9H2O. Show your work.

Answer 1

Answer: The quantity of pure $CaCl_2$ in 7.5 g of $CaCl_2.9H_2O$ is 3.0 g

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}=\frac{7.5g}{273g/mol}=0.027moles$

As 1 mole of $CaCl_2.9H_2O$ contains = 1 mole of $CaCl_2$

Thus 0.027 moles of $CaCl_2.9H_2O$ contains = $\frac{1}{1}\times 0.027=0.027$ mole of $CaCl_2$

Mass of $CaCl_2=0.027mol\times 111g/mol=3.0g$

Thus the quantity of pure $CaCl_2$ in 7.5 g of $CaCl_2.9H_2O$ is 3.0 g