Question

The osmotic pressure of blood is 7.65 atm at 37 °C. What mass of glucose (C6H12O6, molar mass = 180.2 g/mol) is needed to prepare 2.50 L of solution for intravenous injection? The osmotic pressure of the glucose solution must equal the osmotic pressure of blood. (R = 0.08206 L×atm/mol×K)

4.17E-3 g

1.14E3 g

0.751 g

11.1 g

136 g

Match the following aqueous solutions with the appropriate letter from the column on the right.

	1.	0.17 m	Ca(NO3)2		A.	Highest boiling point
	2.	0.18 m	FeBr2		B.	Second highest boiling point
	3.	0.16 m	Ba(CH3COO)2		C.	Third highest boiling point
	4.	0.57 m	Ethylene glycol(nonelectrolyte)		D.	Lowest boiling point

Answer 1

The osmotic pressure of blood is 7.65 atm at 37 °C. What mass of glucose (C6H12O6, molar mass = 180.2 g/mol) is needed to prepare 2.50 L of solution for intravenous injection? The osmotic pressure of the glucose solution must equal the osmotic pressure of blood. (R = 0.08206 L×atm/mol×K)

Osmotic pressure= CRT, C= concentration of glucose , R= 0.08206 L.atm/mole.K, T= 37 deg,c =37+273= 310K

moles of glucose= mass/molar mass= M/180.2, concentration of glucose= M/(180.2* 2.5)= 0.00222M

Hence 7.65= 0.00222M*0.08206*310

M= 135.5 gm

2. Boiling point elevation= i*Kb*m, m= molality, i= Van't Hoff factor, m= molality, Kb= boiling point elevation constant =0.51deg.c/molal

Boiling point of solution = boiling point of water + boiling point elevation=100+ boiling point elevation

i= no of ions formed in solution, for Ca(NO3)2 spilits in Ca+2 ion and 2 NO3- ions. hence i=3

i=3 for FeBr2 ( Fe+2 ion and 2Br- ions)

i=3 for Ba(CH3COO)2 ( Ba+2 and 2 CH3COO- ions)

i=1 for non electrolyte like Ethylene glycol

The data on (ikbm) and boiling point calculations are represented in the form of table

hence as per the calculations, boiling point is highest for FeBr2, second is Ca(NO3)2, third is Ba(NO3)2 and last is ethylene glycol