Question

How many milligrams of lithium bromide (MW = 86.845 g/mol) was dissolved to make a 251 mL aqueous solution with an osmotic pressure of 18.21 atm at 37°C?

- report your answer in 4 sig figs
- do not include units

Answer 1

Answer

7800 mg

Explanation

Osmotic pressure π calculated as follows

π = iMRT

where ,

i = Vant' Hoff factor , for lithium bromide it is 2

M = molarity of solute ,

R = gas constant, 0.082057(L atm/mol K)

T = Temperature, 310.15K

So,

18.21atm = 2× M × 0.082057(L atm/mol K)× 310.15K

M = 0.3578M

Molarity = number of moles of solute per liter of solution

0.3578M LiBr = 0.3578 moles of LiBr per liter soltion

Therefore

Number of moles of LiBr required for 251ml solution = (0.3578mol/1000ml)× 251ml = 0.08981mol

Mass of LiBr required = 0.08981mol × 86.845g/mol = 7.7995g = 7800mg