Question

The molality of 1L solution of 80%H2SO4 (w/V) . given that the density of the solution is 1.80g/ml is

Answer 1

Answer:

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Explanation:

Answer 2

Answer: 8.2 molal

Explanation:

Given: 80 gram of $H_2SO_4$ are dissolved in 100 ml of solution.

Thus if 100 ml of solution contain = 80 gram of  $H_2SO_4$

1000 ml of solution will contain =$\frac{80}{100}\times 1000=800$ gram of $H_2SO_4$

Density of solution= 1.80 g/ml

Now we have to calculate the mass of solution.

$Density=\frac{Mass}{Volume}$  

$\text {Mass of solution}=Density\times Volume=1.80\times 1000ml=1800g$

Mass of solution = Mass of solute + Mass of solvent  

Mass of solvent = mass of solution - mass of solute= (1800- 800)g = 1000 g

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

$Molality=\frac{n\times 1000}{W_s}$

where,

n= moles of solute

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{800g}{98}=8.2moles$  

$W_s$ = weight of solvent in g  = 1000g

Now put all the given values in the formula of molarity, we get

$Molality=\frac{8.2moles\times 1000}{1000g}=8.2mole/kg$

Therefore, the molality of solution will be 8.2 mole/kg.