Solution:
Solution
Weight of salt $$=13.4$$ g
Weight of $$H_2O=6.3$$ g
Weight of anhydrous salt $$=13.4-6.3=7.1$$ g
Moles of anhydrous salt $$=\dfrac {7.1}{842}=0.05$$
Moles of $$H_2O=\dfrac {6.3}{18}=0.35$$ mol
$$0.05$$ mol of anhydrous salt gives $$0.35$$ mol of $$H_2O$$.
$$1$$ mol of anhydrous salt will give $$\dfrac {0.35}{0.05}=7$$ mol of water.
