Question

1e. An aqueous solution has a mole fraction of solute of (4.73x10^-2). The density of the solution is (1.1400x10^0) g/mL and the solute has a molar mass of (8.020x10^1) g/mol. What is the Molarity of solute of this solution?

1f. At an unknown temperature a solution made of (7.740x10^0) g of a non-volatile solute dissolved in 100.0 g of water has a vapor pressure of (5.51x10^1) mm Hg. What is the vapor pressure of pure water (in mm Hg) at this unknown temperature? The molar mass of the solute is (5.360x10^1) g/mol.

1g. A solution is made of two volatile solutes: Chemical A (with a pure vapor pressure of 80.0 mm Hg) and Chemical B (with a pure vapor pressure of 100.0 mm Hg). The solution has a total vapor pressure of (9.88x10^1) mm Hg. If the solution is known to be made from 3.00 mol of A, how many moles of B must there be in the solution?

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Answer 1

g X solute - 4.738 1022 0.047 1 y (density of soln ¿e H2o) - 1.142 mil n solute X solute n solufet n H2o If n solute = 0.047 mole then nH20-1-0.047 molar mass of solute = 0.950 moles = 8.020x10' g most 80.20 g mail 0.3, motes met mass of solute = 80.20g X - 0.953 853 notes 3) 1.14 g mit 20.919 +3.76 = 20.91g va =3.76 g molar mass of solution (H20) = 189 mail mass of H20-189 X mo? = 17.154g pidensitym Total mass a 17.154 =) 20.917 1.148 mit - 18.34 mi -0.018L molarity 2 moles of solate voimcy H20 0.018L in litre - 2.6M AL NOT & Please ask one Oued tim at a time. toomuck Oution at a time is against Chess policy, since Ouestion are long, 0.047 mol CScanite Scanner