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

There are 0.429 moles of Cd and 0.858 moles of N in 132.4 g of Cd(NO3)2·4H2O. Also, there are approximately 1.034 x 10^24 molecules of water of hydration in the same amount.

Step-by-step explanation:

To calculate the number of moles of Cd and N in 132.4 g of Cd(NO3)2·4H2O, we first need the molar mass of the compound. The molar mass of Cd(NO3)2·4H2O is obtained by summing the atomic masses of all the atoms in the formula:

• Cd = 112.41 g/mol
• N = 14.01 g/mol (there are 2 N atoms)
• O = 16.00 g/mol (there are 6 in NO3 and 8 in 4H2O)
• H = 1.01 g/mol (there are 8 in 4H2O)

After calculating, the molar mass is 308.47 g/mol. To find the moles of the entire compound, we divide the mass by the molar mass:

Moles of Cd(NO3)2·4H2O = 132.4 g / 308.47 g/mol = 0.429 moles

Since the formula indicates one mole of Cd and two moles of N per mole of the compound, there are 0.429 moles of Cd and 0.858 moles of N in 132.4 g of Cd(NO3)2·4H2O.

For the water of hydration, there are 4 moles of water per mole of the compound. To find the number of water molecules, we multiply the moles of the compound by Avogadro's number (6.022 x 1023) and then by 4:

Number of water molecules = 0.429 moles x 4 x 6.022 x 1023 ≈ 1.034 x 1024 molecules of H2O.

Answer 2

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* 10^(23)` molecules of water.

Step-by-step 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=(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* 0.430) = 0.430mol` of Cd and
`(2* 0.430)=0.86mol` of N

According to the mole concept:

1 mole of a compound contains
`6.022* 10^(23)` number of molecules

So, 0.430 moles of
`Cd(NO_3)_2.4H_2O` will contain =
`(6.022* 10^(23))/(1mol)* 0.430mol=2.59* 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* 10^(23)` molecules of water.