Question

12% solution of an unknown substance is isotonic with a 6% w/w sorbitol solution. What is its approximate molecular weight assuming it does not dissociate (i=1)?

Hint: Try writing an equation which relates the two osmotic pressures.

Select one:

230 g mol-1

18 g mol-1

546 g mol-1

2400 g mol-1

364 g mol-1

4.6 g mol-1

60 g mol-1

Answer 1

Answer:

364 g mol-1

Explanation:

We know osmotic pressure is calculated by the following formula

$\pi =iMRT$

where pi is the osmotic pressure

i is van't hoff factor

M is molar concentration

R is gas constant

T is temperature

It is given that the compounds does not dissociate so i=1

isotonic means both compound has same osmotic pressure

so we equate the pressure of both the solution we will get the relationship between the concentration of the two solutions

$\frac{Grams of Comp 1}{Molecular Mass of compound 1*Volume of solution 1}=\frac{Grams of comp 2}{Molecular weight of comp 2*Volume of solution}$

This is nothing but molarity of two compounds equated

We are given with 12% w/w solution which means 12 g in 100 g of solvent

and 6% w/w of sorbitol. Let consider molecular mass of unknown is y

If we put all this is in the concentration equation. Considering the same volume

$\frac{12}{y}=\frac{6}{182}$

where 182 is the mass of sorbitol

12 is the weight of unknown and 6 is the weight of sorbitol

so we get y = 364 g/mol