Question

5) Calculate the molality of 0.210 mol of KBr dissolved in 0.075kg pure
water?

Answer 1

`\boxed {\boxed {\sf 2.8 \ m }}`

Step-by-step explanation:

The formula for molality is:

`m=(moles \ of \ solute)/(kg \ of \ solvent)`

There are 0.210 moles of KBr and 0.075 kilograms of pure water.

`moles= 0.210 \ mol \\kilograms = 0.075 \ kg`

Substitute the values into the formula.

`m= ( 0.210 \ mol )/(0.075 \ kg)`

Divide.

`m= 2.8 \ mol/kg= 2.8 \ m`

The molality is 2.8 moles per kilogram