Question

​What is the concentration of the solution prepared by dissolving 2.35 g of KBr (M = 119 g/mol) in 250 mL of water?

Answer 1

The concentration of KBr is
`C = 0.07899 \ mol L^(-1)`

Step-by-step explanation:

From the question we are told that

The mass of KBr is
`m_(KBr) = 2.35 \ g`

The molar mass of KBr is
`M_(KBr) = 119 g/mol`

Volume of water is
`V = 250 \ mL = 250 *10^(-3) = 0.250 \ L`

This implies that the volume of the solution is
`V = 250 mL`

The number of moles of KBr is

`n = (m_(KBr))/(M_(KBr))`

Substituting values

`n = (2.35)/(119)`

`n = 0.01975 \ mol`

The concentration of KBr is mathematically represented as

`C = (0.01975)/(0.250)`

`C = 0.07899 \ mol L^(-1)`