Question

Values of molarity (M) and molality (m) for the
resulting H2SO4, solution are:
1.7.6, 7.6
3.11.2, 7.6
2.7.6, 11.2
4.7.6, 9.5
​

Answer 1

Explanation:

ans: Correct option is A) 7.61,11.22

mark it as brilliant

Answer 2

Answer:

for mixture 1

d1 = 1.218 gram/mol

let volume be v ml

mass = m1 = 1.218 v gram

now 30% by weight means 30 gram H²SO⁴ in 100 gram of solution

100 gram of solution = 30 gram H²SO⁴

1 gram solution = (30/100)×1.218 => 0.3654 v gram

for mixture 2

d2 = 1.6 gram/mol

let volume be v ml

mass = m2 = 1.61 v gram

now 70% by weight means 70 gram H²SO⁴ in 100 gram of solution

100 gram of solution = 70 gram H²SO⁴

1.61 v gram solution = (70/100)×1.61 v gram => 1.127 v gram

hence total mass of H²SO⁴ = (0.3654 v gram + 1.127 v gram) => 1.4924 v gram

moles of H²SO⁴ = given weight/ molecular weight

moles of H²SO⁴ = 1.4924 v / 98 => 0.0152 v moles

total volume of solution = (v+v) ml => 2v ml

Molarity (M) = no. of moles of H²SO⁴ / total volume

Molarity (M) = 0.0152 ml/ [(2v/1000) litre] => 7.60M

Total mass of solution = m1+m2 => (1.218v + 1.610v) gram => 2.828 v gram

total mass of solute = 1.4924v gram

total mass of solvent = (2.828 v - 1.4924 v) gram => 1.3356 v gram

moles od H²SO⁴ = moles of solute = 0.0152 v mol

Molality (m) = moles of solute / mass of solvent (in kg)

Molality (m) = 0.0152/ [(1.3356/1000)v]

Molality (m) = 11.22

hence molarity = 7.60M and molality = 11.22

Explanation:

please mark this answer as brainliest