the molality of 1L solution with x% H2SO4 is 9 The weight of solvent present in the solution - Chemistry - Equilibrium

Question

the molality of 1L solution with x% H2SO4 is
9 The weight of solvent present in the solution is 910g the value
of x is?
a 90
b 80 3
c 30 38
d 46 87

Vartika Jain · Accepted Answer

Dear Student,

Molality =Number of molesMass of solvent ( in kg)or, Number of moles = Molality×Mass of solvent (in kg)                                     = 9×0
910                                      =8
19 molNow, Mass of H2SO4=Molar mass of H2SO4×Number of moles = 98×8
19 = 802
62 g% weight of H2SO4= x =Mass of H2SO4Mass of solution=Mass of H2SO4Mass of H2SO4+Mass solvent=0
8020 802+0 910=0 8021 712=46
87Hence, correct answer is (d)

the molality of 1L solution with x% H2SO4 is 9. The weight of solvent present in the solution is 910g. the value of x is? a. 90 b. 80.3 c. 30.38 d. 46.87

the molality of 1L solution with x% H₂SO₄is 9. The weight of solvent present in the solution is 910g. the value of x is?
a. 90
b. 80.3
c. 30.38
d. 46.87