Question

4) A solution is prepared by dissolving 15.2 g of NaC,H,O, into water. The concentration of the solution is measured at 1.95 M. Calculate the volume of the solution (in Liters): (5) (MM NaC,H,O,: 82.03 g/mol) 5) Common intravenous (IV) glucose solutions are prepared as 5.0 % mlv glucose. How many mililiters of solution are present in an IV bag that contains 39 grams of glucose? 6) Two solutions are prepared. The first contains glucose (C.H.O.) dissolved in water at 0.180 M concentration. The second contains 13.0 grams of Mgci, in water, with a total volume of 1.77 Liters. (MgCl, = 95.21 g/mol) a) Calculate the Molarity concentration of of the MgCl, solution: (4) b) Calculate the osmolarity concentration of both solutions: (2) Which solution would you predict to have the highest boiling point?

Answer 1

Ans

4. Mass of sodium acetate taken = 15.2g

Molar mass of sodium acetate=82.03 g/mol

Molarity of the solution = 1.95 M

We know, molarity of a solution = no.of moles of the solute/volume of thesolution

   Or, Volume of the solution=no.of moles/molarity

=15.2/82.03$\ast$1.95

=0.095 L

5. Concentration of intravenous glucose solution = 0.5% m/v

= 0.5 gm glucose in 100 ml solution

volume of soltion which contain 39g of glucose=100$\ast$39/0.5

=7800ml

6.a) Mass of magnesium chloride =13 g

molar mass of magnesium chloride= 95.21g/mol

Volume of solution    = 1.77 L

Molarity of MgCl₂ solution =no.of moles of solute/volume of solution

= 13/95.21$\ast$1.77

=0.0771M

c) Molarity of glucose =0.18 M

Molarity of MgCl₂ solution=0.0771 M

Since the concentration of glucose solution is more,its boiling point elevation will also be more. Hence its boiling point will be more.

b) Osmolarity of a solution is the total number of solute particles per litre

Osmolarity of glucose solution=0.18$\ast$ 6.022$\ast$10²³

Osmolarity of magnesium chloride solution = 0.0771$\ast$6.022$\ast$10²³