Question

How many moles of Cd and of N are contained in 132.4 g of Cd(N03)2-4H20? (b) How many molecules of water of hydration are in this same amount?

Answer 1

Answer: The given amount of $Cd(NO_3)_2.4H_2O$ contains 0.430 moles of Cd, 0.860 moles of N and $2.59\times 10^{23}$ molecules of water.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $Cd(NO_3)_2.4H_2O$ = 132.4 g

Molar mass of $Cd(NO_3)_2.4H_2O$ = 308.5 g/mol

Plugging values in equation 1:

$\text{Moles of }Cd(NO_3)_2.4H_2O=\frac{132.4g}{308.5g/mol}=0.430 mol$

1 mole of $Cd(NO_3)_2.4H_2O$ contains 1 mole of Cd, 2 moles of nitrogen atom (N), 10 moles of oxygen atom (O) and 8 moles of hydrogen atom (H).

So, 0.430 moles of $Cd(NO_3)_2.4H_2O$ will contain = $(1\times 0.430) = 0.430mol$ of Cd and $(2\times 0.430)=0.86mol$ of N

According to the mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, 0.430 moles of $Cd(NO_3)_2.4H_2O$ will contain = $\frac{6.022\times 10^{23}}{1mol}\times 0.430mol=2.59\times 10^{23}$ number of water molecules.

Hence, the given amount of $Cd(NO_3)_2.4H_2O$ contains 0.430 moles of Cd, 0.860 moles of N and $2.59\times 10^{23}$ molecules of water.