# What is the molality of a solution in which 2.3 g O_2 gas dissolved in 355 g of water?

Jun 11, 2016

$0.20 m$

#### Explanation:

Molality can be found using the equation below:

What you want to do now is determine what the solute and solvent are. A solute is dissolved in the solvent and the solute is the "minor component in a solution."The solvent is usually water or it is the substance that you have more of. In our case, the solute is ${O}_{2}$ and the solvent is water.

As we can see from the equation, the solute and solvent have the wrong units. Let's start by converting grams of ${O}_{2}$ to moles of ${O}_{2}$. We do this by using the molar mass of ${O}_{2}$ as a conversion factor.

2.3cancel"g"O_2xx (1mol O_2)/(32.00cancel"g")= 0.0719 mol O_2

Now we need kg of solvent instead of the grams that we're given. Let's use this relationship:

355cancel"g" H_2O xx (1kg)/(1000cancel"g") = .355kg H_2O

Now all we do is plug the values into the equation like this:

$m o l a l i t y = \frac{0.0719 m o l}{0.355 k g}$ = $0.20 m$