Question

the molarity of H2SO4 solution, which has a density 1.84g/cc at 35° C contains 98 % by weight is

Answer 1

as the given H2SO4 is 98%by wt, hence
100g solution contains 98g H2SO4 by mass
now density d=(m/v) ∴v=(m/d)
here, d=1.84g/c.c. m=100g, v=?
∴v=(100/1.84)mL=$\frac{100}{1.84*1000} L$
 again molarity=$\frac{W}{M*V}$
here weight taken W=98g, molecular weight of H2SO4 (M)=98g/mol,
volume (V)=$\frac{100}{1.84*1000} L$
putting these values we get
∴molarity=(98*1.84*1000)/(98*100)=18.4(M)

Answer 2

Answer:18.41..

Explanation:

the solution contains 98% H2SO4 by weight/mass

that means 100 gram of solutions contains 98 grams of H2SO4

given its density is 1.84 gm/ml (cc = ml)

we know density = mass/volume

so volume of the solution = mass of solution / density of solution

= 100 gram / 1.84 = 54.3 ml

No. of moles of H2SO4 = weight of H2SO4 given / molar mass of H2SO4

= 98/98 = 1 mole

so molarity = 1 mole / 54.3 ml * 1000 = 18.41 M