Question

0.5 M urea is isotonic with solution of 10g non volatile , non electrolyte unknown solute X in 100ml solution. The molar mass of the solute is​

Answer 1

Given:

The urea solution and unknown solution are isotonic.

The molarity of urea solution, M1 = 0.5 M

The mass of unknown solute, w = 10 gm

The volume of the unknown solution, V = 100 ml

To Find:

The molar mass of the unknown solute.

Calculation:

The molarity of the unknown solution is given as:

M2 = (w×1000)/(M.wt×V)

⇒ M2 = (10×1000)/(M.wt×100)

⇒ M2 = 100/M.wt

- As the urea solution and unknown solution are isotonic, it means their osmotic pressure is equal.

⇒ Π(urea) = Π(unknown)

⇒ M1RT = M2RT

⇒ M1 = M2

⇒ 0.5 = 100/M.wt

⇒ M.wt = 100/0.5

⇒ M.wt = 200 gm/mol

- So, the molar mass of the unknown solute is 200 gm/mol.

Answer 2

Given:

Concentration of solution of urea, $C=0.5M$

mass of non volatile solution, $m=10g$

Volume of solution, $V=100ml$

To Find:

We have to find the molar mass of solute.

Solution:

By considering the given data,

Osmotic pressure of urea is same as the osmotic pressure of solute,

CRT of urea = CRT of solute

and we know that  $C=\frac{n}{V}$

now,

$0.5RT=\frac{moles}{100/1000} *RT$

$moles=\frac{50}{1000}$

We know that, moles = given mass/ molar mass

$molar mass=\frac{10000}{50}=200g/mol$

Hence the molar mass of solute is $200g/mol$

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