# 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?

Here's how you can solve this problem.

Use the molecular mass of the acid to do that.

This indicates that the molarity of the solution is

Molality can be defined as the mass of the solvent (in kilograms) divided by the moles of solute (which you have already calculated)!

Use the density of the solution to find the total weight of the solution in order to calculate the mass of water you have.

This implies that the water's mass will be

Thus, the solution's molality will be

The amount of protons that sulfuric acid will produce in solution will be taken into account when determining the normality of the solution.

You will receive twice as many equivalents because you have 0.11 moles.

Accordingly, the normality of the solution will be twice as large as its molarity.

You must first ascertain how many moles of water you have in order to obtain the mole fraction of the solution.

There will be a total of moles of

Sulfuric acid's mole fraction will therefore be

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Molarity: 1.1 M Molality: 1.08 m Normality: 2.2 N Mole fraction: 0.04

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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