Question

What mass of silver ag which has an atomic mass of 107.87 amu contains the same number of atoms contained in 10.0g of boron b which has an atomic mass of 10.81 amu?

Answer 1

To contain the same number of atoms also mean to contain the same number of moles. So let us say that X is the mass of Silver Ag required, so that:

X / 107.87 = 10 / 10.81

X = 99.79 g

Answer 2

Answer: The mass of silver atoms is 99.78 grams.

Explanation:

To calculate the number of moles, we use the equation:

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

For boron:

Given mass of boron = 10.0 g

Molar mass of boron = 10.81 g/mol

Putting values in equation 1, we get:

$\text{Moles of boron}=\frac{10.0g}{10.81g/mol}=0.925mol$

As, the number of atoms of silver are same as the number of atoms of boron. This means that number of moles of boron will be same as number of moles of silver.

Now, calculating the mass of silver atoms by using equation 1, we get:

Molar mass of silver = 107.87 g/mol

Moles of silver = 0.925 moles

Putting values in equation 1, we get:

$0.925mol=\frac{\text{Mass of silver}}{107.87g/mol}\\\\\text{Mass of silver}=99.78g$

Hence, the mass of silver atoms is 99.78 grams.