Q:- What is the molality of 98% H_{2}SO_{4} by weight?

An approach to solution:- Let we took 100 g of H_{2}SO_{4} solution. It means that we have 98 g H_{2}SO_{4} in the solution. Let us assume the solvent to be Water (H_{2}O). So the water comes out to be 2 g.

Wt. of solute (H_{2}SO_{4}) = 98 g

Mol. wt. of solute (H_{2}SO_{4}) = 98 g

Wt. of solvent(H_{2}O) = 2 g

Now according to formula of molality:-

(98*1000) / (98*2)= 500

But it is incorrect...

Again let us go to the definition of solute and solvent:

Solute: It is present is smaller amount, and

Solvent: It is present is greater amount.

But here we assumed the H_{2}SO_{4} to be solute which is 98 g and H_{2}O to be solvent which is 2 g.

Let us find the solution by assuming H_{2}SO_{4} as **solvent** and H_{2}O as **solute**.

Wt. of solute(H_{2}O) = 2 g

Mol. wt. of solute(H_{2}O) = 18 g

Wt. of solvent(H_{2}SO_{4}) = 98 g

According to formula: (2*1000) / (18*98) = 1.13

Both the answers are wrong as the answer is 10 m.

If in first case

Wt. of solute (H_{2}SO_{4}) = 98 g

Mol. wt. of solute (H_{2}SO_{4}) = 98 g

**Wt. of solvent(H**_{2}O) = 2 g, we change it to

**Wt. of solution = 100 g**

(98*1000) / (98*100) = 10 m but How??

Where am i wrong??please tell me...