Question

@Abmon98

Answer 1

The molarity of 98% \(\sf H_2SO_4\) (d=1.8g/ml) by weight is ?

Answer 2

Answer is 18 M,
But how ?

Answer 3

My attempt

Let the mass of H2SO4 be n and that of water be m

\(\sf \huge \frac{n}{n+m}=0.98\)

Now we just need to convert the denominator into volume

Answer 4

so if i divide 0.98 by 1.8 i get 0.54....but that's not the answer

Answer 5

Molarity of a solution is defined as the number of moles of the solute dissolved per litre (or dm3) of solution. Molarity=moles of solute/Volume of solution(L)
M = Volume of the solution's in litre /Number of moles of solute
98% solution of H2SO4 means 98g of H2SO4 arepresent in 100g of the solution
Wt. of H2SO4 dissolved = 98 g
Weight of the solution = 100 g
Density of the solution = 1.8 g/ml
Molecular weight of H2SO4=98
Number of mole=Mass(g)/Molar Mass(g/mol)=98/98=1 mole
thus the 2 grams of solvent contain 1 mole of solute,
Volume of solution=weight of solution/Density of solution
=100/1.8=55.55556 ml
This 55.5555556 cm3 of solution contain H2SO4
= 1 mole
1000 cm3 of solution contain H2SO4
1000*1/55.55556=17.9999

Answer 6

Thank you very much for your time and work...appreciate it !

Answer 7

your most welcome :D

Answer 8

is there any significance of this line ?
`thus the 2 grams of solvent contain 1 mole of solute,`

Answer 9

i made a mistake here they both share the same number of moles ,it wouldnt change unless an extra volume was added.

Answer 10

yeah..i got it !
Thanx :)

Answer 11

good to hear that :), dont mention it :D.

Answer 12

\(\huge\sf\color{green}{\text{✌ﾟ\(\ddot\smile\) ✌ﾟ}}\)