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

Answer: The molality of the solution is 0.1 m.

Explanation:

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

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

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:

$\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m$

Hence, the molality of the solution is 0.1 m.

Answer 2

Answer:

the molality of the solution is 10 mol/kg

Explanation:

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

$m=\frac{moles of solute}{kg of solvent}$

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

$1000 g \longrightarrow 1Kg\\100g \longrightarrow x \\x=\frac{(100)(1)}{1000}\\ x=0.1 Kg$

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

$m=\frac{moles of solute}{kg of solvent}\\m=\frac{1mol}{0.1kg}\\ m=10 \frac{mol}{kg}$

the molality of the solution is 10 mol/kg