Question

how many grams of solute is required for the preparation of 3.50 molar solution of sulphuric acid.the volume of the solution is 500ml.

Answer 1

how many grams of solute is required for the preparation of 3.50 molar solution of sulphuric acid. the volume of the solution is 500ml.

In chemistry, the molar concentration, $c_{i}$ is defined as the amount of a constituent $n_{i}$ (usually measured in moles - hence the name) divided by the volume of the mixture $V$:

$$c_{i} = \frac{n_{i}}{V}$$

It is also called molarity, amount-of-substance concentration, amount concentration, substance concentration, or simply concentration. The volume $V$ in the definition $c_{i} = n_{i} / V$ refers to the volume of the solution, not the volume of the solvent. One litre of a solution usually contains either slightly more or slightly less than 1 litre of solvent because the process of dissolution causes volume of liquid to increase or decrease. So if molarity is $3.5\mathrm{M}$ and volume is $0.5\mathrm{L}$, the amount of acid is:

$$\mathrm{n} = \mathrm{C} * \mathrm{V} = 0.5 * 3.5 = 1.75 \text{ moles}$$

The mass of acid can be found from next:

$$\mathrm{n} = \mathrm{m} / \mathrm{Mw}, \text{ where } \mathrm{Mw} \text{ is molecular mass}.$$ $$\mathrm{m} = \mathrm{n} * \mathrm{Mw} = 1.75 * 98 = 171,5 \mathrm{g}$$