What would be the molality of a solution obtained by mixing equal volumes of 30% by weight of H2SO4(d=1.218 g/ml) and 70% by wt of H2SO4(d=1.61 g/ml)?

Assume y ml of each solution, a and b, are mixedCalculate mass of solution using formula, mass = density(d) × Volume (V)Mass of solution a, ma = 1.218 × y = 1.218yMass H2SO4, msolute a = 30% or 30×1.218y100 msolute a = 0.3654y g Moles H2SO4, nsolute a = 0;3654y98or 0.003729y mol (Using formula, moles = mass, m /Molar mass) Mass of water, msolvent a = mA- msolute a msolvent a = 1.281y – 0.3654y msolvent a= = 0.8526y g or 0.8526 X 10-3y kg Similarly for solution B mb = 1.610y g msolute b = 70×1.610y100or 1.127y g nsolute b = 1.127y98 or 11.5y mol msolvent b = 1.610y – 1.127y = 0.483y g or 0.483 × 10-3y kg When y ml of each solution is mixedTotal volume, V = y + y = 2y ml or 2y × 10-3 LTotal moles of H2SO4, nsolute = nsolute a + nsolute b nsolute= 0.003729y + 11.5y = 0.015229y mol Total mass of solvent in kg, msolvent = msolvent a + msolvent bmsolvent = 1.3356y×10-3 kgMolarity, M =nsolute/VM = 0.015229y2y×10−3 M = 7.614MMolality, m = nsolutemsolvent in kg=0.015229y1.3356y×10−3=11.402m