Question

What volume of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?

Answer 1

We know that

molecular mass of Sulphuric Acid = [math](2 * 1)[/math] [math]+[/math] [math](1 * 32)[/math] [math]+[/math] [math](4 * 16) [/math] = [math]98 u .[/math]

Now ,

The number of moles of [math]H_2 SO_4 [/math] in 2.45 grams is

n = [math]\frac {2.45}{98} [/math] = [math]0.025 mol .[/math]

Now the concentration is

[math][H_2 SO_4] [/math] [math]=[/math] [math]18 [/math] [math]Molar [/math] = [math]18 [/math] [math]\frac {moles}{litre}[/math]

Since

18 moles are present in 1 litre ,

1 mole is present in [math]\frac {1}{18} [/math] litres , and

0.025 moles are present in 0.025 [math] * \frac {1}{18} [/math]litres

0.025 moles are present in 0.001389 litres = 1.389 mL .

Thus , 1.389 mL of 18 M Sulphuric Acid contain 2.45 grams Sulphuric Acid .

its preety easy tbh

Answer 2

1.39 ml of 18.0 M $H_2SO_4$ is needed to contain 2.45 g  

Explanation:

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute

Moles=

= volume of solution in ml= ?

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

$18.0M=\frac{0.025moles\times 1000}{V_s}$

$V_s=1.39ml$

Learn more about molarity

https://brainly.com/question/9162756

https://brainly.com/question/9917809