Calculate the molarity of a 588 m aqueous ethanol CH3CH2OH s

Calculate the molarity of a 5.88 m aqueous ethanol (CH₃CH₂OH) solution with a density of 0.930 g/mL.

A.4.99 M

The osmotic pressure of human blood is 7.6 atm at body temperature, 37

Solution

molality = moles of solute/mass(Kg) of solvent = 5.88m

hence mass of CH₃CH₂OH = 5.88*46 = 270.48gm

gence total mass = 270.48+1000 = 1270.48g

hence volume = mass/density = 1270.48/0.93= 1366.107ml = 1.366L

so molarity = 5.88/1.366 = 4.305M........................ANS(D)

T = 37C = 310K

P = 7.6atm = 7.6*11325 Pa

R = 8.314J/mol-k

P = MRT

M = P/RT= 298.78 moles/m^3 = 0.299 M

so concentration of glucose = 0.299 M

for 1L moles og glucose = 0.299 moles

molar mass of glucose = 180g

mass of glucose required = 180*0.299 = 53.82gm = 54 gm..............ANS(B)