Question

How many grams of calcium lactate pentahydrate would be recovered from $1$ g of anhydrous salt?

• A. $1.41$
• B. $1.00$
• C. $1.27$
• D. $1.51$

Answer 1

The correct option is A $1.41$

As we know, molecular weight of anhydrous calcium lactate $=218$ g/mol.
Number of moles in $1$ g anhydrous salt $=\frac{1}{218}=4.587\times {10}^{-3}$.
$1$ mol pentahydrate salt would be recovered from $1$ mol anhydrous salt.
Thus, number of moles of pentahydrate salt $=4.587\times {10}^{-3}$.
Molecular weight of pentahydrate salt $=$ Molecular weight of anhydrous salt $+$ Weight of $5$ water molecules $=218+18\times 5=308$ g/mol
Thus, mass of pentahydrate salt $=308\times (4.587\times {10}^{-3})=1.41$ g.