Question

If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the resulting solution?

Answer 1

the molality of the solution is 10 mol/kg

Step-by-step explanation:

To find the molality of a solution we apply the following formula

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

data:

100 g of solvente

We have the amount of solvent expressed in grams but to be able to use the data in the formula we must pass it to kilograms

we apply a simple rule of three

`1000 g \longrightarrow 1Kg\\100g \longrightarrow x \\x=((100)(1))/(1000)\\ x=0.1 Kg`

27.1g of solute, molar mass 27.1 g/mol

We need the moles of solute and they give us grams of solute and molar mass

The molar mass is 27.1 g / mol This means that there is 27.1 g for each mol of solute.

The amount of solute we have is 27.1g and according to the molar mass this amount of solute corresponds exactly to 1 mole

Therefore having 27.1 g of solute is equivalent to saying that we have 1 mol of solute

Now that we have the data in the correct units we apply the formula

`m=(moles of solute)/(kg of solvent)\\m=(1mol)/(0.1kg)\\ m=10 (mol)/(kg)`

the molality of the solution is 10 mol/kg

Answer 2

Answer: The molality of the solution is 0.1 m.

Step-by-step explanation:

To calculate the molality of solution, we use the equation:

`Molality=\frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ in grams}}`

Where,

`m_(solute)` = Given mass of solute = 27.1 g

`M_(solute)` = Molar mass of solute = 27.1 g/mol

`W_(solvent)` = Mass of solvent = 100 g

Putting values in above equation, we get:

`\text{Molality of solution}=(27.1* 1000)/(27.1* 100)\\\\\text{Molality of solution}=0.1m`

Hence, the molality of the solution is 0.1 m.