Molality is a measure of concentration defined as moles of solute per kg of solvent. So if the molality of the sulfuric acid solution is 9, then that means that there are 9 moles of sulfuric acid per 1 kg of solvent. So we have a solution that is 9 molal in concentration and the amount of solvent in it is 910 g which is the same as 0.910 kg. So we multiply the two to get the moles of sulfuric acid in the solution.
(9 mole / kg) * 0.910 kg = 8.19 moles sulfuric acid
Multiply this by the molecular weight of sulfuric acid (98.079 g/mole) to convert into grams.
8.19 moles * (98.079 g / mole) = 803.3 g sulfuric acid
Now divide by the density of sulfuric acid (1.84 g/mL) to convert the grams into mL's.
803.3 g * (1 mL / 1.84 g) = 436.6 mL sulfuric acid
So we have 436.6 mL of sulfuric acid total. Since we know that we have 1 L (the same as 1,000 mL) of total solution, we know that means that the percentage of the sulfuric acid in the solution can be calculated as shown below.
436.6 mL H2SO4 / 1,000 mL total solution = 0.437 = 43.7%
So we know that the value of x in the original question (percentage of H2SO4 in the solution) is 43.7%
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