Question

What is the molarity of a solution containing 12.25 g of NaBr in 150 mL of solution?

Answer 1

The molarity is
`M = 0.7937 \ mol/L`

Step-by-step explanation:

From the question we are told that

The mass of NaBr
`m_b = 12.25 \ g`

The volume of the solution is
`V = 150 m L = 150 *10^(-3 )L`

The number of moles of NaBr is

`n_b = (m_b)/(M_b )`

Where
`M_b` is the molar mass of NaBr which is a constant with value
`M_b = 102.894 g/mol`

So

`n_b = (12.25)/(102.894)`

`n_b = 0.1191 \ mol`

The Molarity is mathematically evaluated as

`M = (n_b)/(V)`

Substituting values

`M = (0.1191)/(150 *10^(-3))`

`M = 0.7937 \ mol/L`