Question

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Answer 1

Given a density 1.122 g/mL and a H2SO4 molality of 4.500 m, find the molarity, mole fraction and mass percent.

Solution:

1) The given molality means 4.500 mol dissolved in 1.000 kg of water. Determine the mass of each component:

H2SO4 ---> (4.500 mol) (98.078 g/mol) = 441.351 g 
H2O ---> 1.000 kg = 1000. g

2) Determine mass percentages:

1000. g + 441.351 g = 1441.351 g (total mass of the solution)

H2O ---> 1000. g / 1441.351 g = 69.38% 
H2SO4 ---> 100 - 69.38 = 30.62%

3) Determine mole fraction:

H2SO4 ---> 4.500 mol 
H2O ---> 1000. g / 18.015 g/mol = 55.509 mol

55.509 mol + 4.500 mol = 60.009 mol

H2SO4 ---> 1 - 0.9250 = 0.0750 
H2O ---> 55.509 mol / 60.009 mol = 0.9250

4) Determine the molarity:

1441.351 g / 1.122 g/mL = 1296.179 mL = 1.296179 L

4.500 mol / 1.296179 L = 3.472 M

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Answer 2

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