Question

1. An old sample of concentrated sulfuric acid to be used in the laboratory is approximately
98.4 percent H₂SO₄ by mass. Calculate the molality and molarity of the acid solution. The density of the solution is 1.83 g/mL.

____ m

___  M

2. Calculate the vapor pressure of a solution made by dissolving 255 g of urea[(NH₂)₂CO; molar mass 60.06 g/mol] in 495 g of water at 25°C.

_____ mm Hg   (At 25°C, P_H2O = 23.8 mmHg)

Answer 1

1)
A)
finding molality:
Let mass of solution be 1 Kg = 1000 g

mass of H2SO4 = 98.4 % of mass of solution
= 98.4*1000/100
= 984 g
mass of solvent = mass of solution - mass of solute
mass of solvent = 1000 g - 984 g
mass of solvent = 16.0 g
mass of solvent = 0.016 Kg

Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol


mass(H2SO4)= 984.0 g

use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(984.0 g)/(98.086 g/mol)
= 10.03 mol

m(solvent)= 0.016 Kg

use:
Molality,
m = number of mol / mass of solvent in Kg
=(10.03 mol)/(0.016 Kg)
= 627 molal

B)

finding molarity:
Let volume of solution be 1 L
volume , V = 1 L
= 1*10^3 mL


density, d = 1.83 g/mL
use:
mass = density * volume
= 1.83 g/mL *1*10^3 mL
= 1830.0 g
This is mass of solution
mass of H2SO4 = 98.4 % of mass of solution
= 98.4*1830.0/100
= 1800.72 g

Molar mass of H2SO4,
MM = 2*MM(H) + 1*MM(S) + 4*MM(O)
= 2*1.008 + 1*32.07 + 4*16.0
= 98.086 g/mol


mass(H2SO4)= 1800.72 g

use:
number of mol of H2SO4,
n = mass of H2SO4/molar mass of H2SO4
=(1800.72 g)/(98.086 g/mol)
= 18.36 mol
volume , V = 1 L


use:
Molarity,
M = number of mol / volume in L
= 18.36/1
= 18.4 M
Answer:
627 m
18.4 M

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