Question

If 6.84% (weight/volume) solution of cane sugar (molecular weight = 342 g/mole) is isotonic with 1.52% (weight /volume) solution of thiocarbamide, then the molecular weight (g/mole) of thiocarbamide is:
a.) 152
b.) 76
c.) 260
d.) 180

Answer 1

Hint: Attempt this question by equating the osmotic pressure of both the solutions as both the solutions are isotonic with each other. Remember that osmotic pressure = molar concentration of dissolved species \[\times \] ideal gas constant (R) $\times $ temperature of the solution (T).

Complete answer:
Given data:
As 6.84% (weight/volume) cane sugar solution is prepared. Therefore, the weight of cane sugar in 100 mL of solvent is 6.84 g.
Similarly, 1.52% (weight/volume) thiocarbamide solution consists of 1.52 g of thiocarbamide in 100 ml of solvent.
Therefore given data are:
Weight of cane sugar (\[w1\]) = 6.84 g
Volume of cane sugar solution ($v1$) = 100 mL
Molecular weight of cane sugar ($M1$) = 342 g/mole
Weight of thiocarbamide ($w2$) = 1.52 g
Volume of thiocarbamide solution ($v2$) = 100 mL
Molecular weight of thiocarbamide ($M2$) = ? g/mole
It is also stated that cane sugar solution is isotonic with thiocarbamide solution. For solutions to be isotonic, their osmotic pressure should be equal.
Therefore, osmotic pressure of cane sugar solution = osmotic pressure of thiocarbamide solution.
As, osmotic pressure = molar concentration of dissolved species \[\times \] ideal gas constant (R) $\times $ temperature of the solution (T)

Therefore, on equating the osmotic pressures of both the solutions:
\[C1\times R\times T=C2\times R\times T\]
As R and T are the same on both sides, we will cancel R and T with each other. Therefore, we are left with concentrations on both sides (L.H.S. and R.H.S.)
\[C1=C2\]

Now, putting the value of concentration, which is equal to $\dfrac{w\times 1000}{M\times V}$, we get
$\dfrac{w1\times 1000}{M1\times V1}=\dfrac{w2\times 1000}{M2\times V2}$
Now, we will place all the given data values,
\[\dfrac{6.84\times 1000}{342\times 100}=\dfrac{1.52\times 1000}{M2\times 100}\]
\[M2=\dfrac{1.52\times 1000\times 342\times 100}{100\times 6.84\times 1000}\]
\[M2=\dfrac{1.52\times 342}{6.84}\]
\[M2=76\]g/mole
Therefore, the molecular weight of thiocarbamide is 76 g/mole.
So, the correct answer is “Option B”.

Additional Information: Isotonic solutions are those solutions which have the same osmolarities or solute concentrations. If such solutions are separated via a semipermeable membrane then water will pass from equal part of each solution into another solution. Isotonic condition allows water to pass from one solution to another without altering the solute concentration.

Note: In an MCQ type question, if (weight/volume) of isotonic solution is given then directly apply,
$\dfrac{w1}{M1}=\dfrac{w2}{M2}$