# An aqueous solution of #H_2SO_4# (molecular weight=98) contains 10.78 #g# of acid per #dm^3# of solution. Density of solution is 1.123 #g#/#ml#. What is the molarity, molality, normality and mole fraction of the solution?

##### 1 Answer

Here's how you can solve this problem.

#### Explanation:

The problem actually makes your life easier by telling you that you have that much sulfuric acid, **10.78 g** to be precise, per

If you take into account that fact that **10.78 g**.

To do that, use the acid's moalr mass

This means that the solution's molarity is

Molality is defined as moles of solute, which you've already calculated, divided by the mass of the solvent - expressed in **kilograms**!

To get the mass of water you have, use the solution's density to first determine how much the entire solutions weighs.

This means that the mass of water will be

This molality of the solution will thus be

The solution's **normality** will take into account the number of protons the sulfuric acid will produce in solution.

Since sulfuric acid is a *diprotic acid*, every mole of the acid will produce **two moles** of **equivalents**.

Since you have 0.11 moles, you'll get twice as many equivalents.

This means that the solution's normality will be **twice as big** as its molarity.

To get the *mole fraction* of the solution, you need to first determine how many moles of water you have.

The *total number of moles* will be

The *mole fraction* of sulfuric acid will thus be