Question

1) A 6.01 E-5 M solution of sugar water at 25C has what osmotic pressure? (a)4.33E-4 atm (b)2.33E-3atm (c)2.92E-3 atm (d)1.47E-3 atm

2) 10.5g of Potassium Bromide(MM 119) is dissolved in 45.0 g of water. If the Kb of water is 0.512C/m, what is the boiling point? (a)98C (b)102C (c)101C (d)105C

3) Vitamin D is a fat-soluble vitamin. Which of the following solvents will dissolve it? (a) CH3CH2O-H (b)H2O (c)CH3CN (d)Hexanes -CH3CH2CH2CH2CH2CH3

4) How many grams of glucose (C6H12O6) need to be added to 4.50L of water to lower the pressure to 75.0% of its original pressure. (a)1.35E6 g (b)1.49E4 g (c)6.00E4 g (d)83 g

Answer 1

1) Osmotic pressure = cRT

c = 6.01×10^-5 M, T = 25°C = 25+273.15 K = 298.15 K

R = 0.0821 Latm/mol.K

Osmotic pressure = 6.01×10^-5×0.0821×298.15 atm

= 1.47×10^-3 atm

Answer is d).

2) Boiling point = Tb° + iKbm

Van't Hoff factor, i = 2 (because two ions are formed by KBr)

Molality, m = mass of KBr×1000/molar mass×mass of water

= 10.5×1000/119×45 = 1.96 m

Tb° = 100°C

Thus, boiling point = 100 + 2×0.512×1.96 = 102°C

Answer is b).

3) Answer is d). Fat are nonpolar thus nonpolar solvent dissolve vitamin D which is hexane.

4) Pressure lowered, ∆P = 0.25×P

Where, P is initial pressure

Moles of water = mass/molar mass

= 4500/18 = 250 g

(4.5L = 4500 ml = 4500 g because water has density of 1)

Thus, ∆P/P = m/180×250

0.25P/P = m/180×250

m = 25×180×250/100

= 1.49×10⁴ g

Answer is b).