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

Answer:

The concentration of KBr is  $C = 0.07899 \ mol L^{-1}$

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

         $n = \frac{m_{KBr}}{M_{KBr}}$

Substituting values

         $n = \frac{2.35}{119}$

        $n = 0.01975 \ mol$

The concentration of KBr is mathematically represented as

                $C = \frac{0.01975}{0.250}$

                $C = 0.07899 \ mol L^{-1}$