Question

A 4% solution of Cane sugar C12H22O11 is isotonic with 3% solution of An unknown organic compound calculate the mol mass of unknown organic compound​

Answer 1

Answer:

For easy calculations let us consider that the mass of solution is 100 g.

Therefore,  

Mass of 4% solution of cane sugar is = 4 g

Mass of 3% solution of unknown organic compound is = 3 g

The molecular mass of cane sugar = 342.29 g/mol

Let the molecular mass of the unknown compound be “x” g/mol

Also, given that both the solutions are isotonic so we can consider that their molar concentration is equal.

Therefore,

[Molar concentration of cane sugar] = [Molar concentration of unknown compound ]

⇒ [moles of cane sugar * 1litre of solution] = [moles of unknown compound * 1 litre of solution ]

⇒ (4/342.29) * (1000/100) = (3/x) * (1000/100)

⇒ x = (3 * 342.29) / 4

⇒ x = 256.71 g/mol

Thus, the molecular mass of the unknown organic compound is 256.71 g/mol .

Answer 2

Answer:

Explanation:

At first , let the mass of the solution be  100 g.

So,

Mass of 4% solution of cane sugar is = 4 % of 100 = 4g

Mass of 3% solution of unknown organic compound is = 3 % of 100 = 3g

Since,  The molecular mass of cane sugar = 342.29 g/mol

Let the molecular mass of an unknown compound be “x” g/mol

Since , the solution has the same osmotic pressure;

Also, it is being given that both the solutions are isotonic in nature,

So we will consider that their molar concentration is equal.

Therefore,

[Molar concentration of cane sugar] = [Molar concentration of unknown compound ]

$[moles of cane sugar \times  1 litre of solution] = [moles of unknown compound \times 1 litre of solution ]$

⇒  $(4/342.29) \times (1000/100) = (3/x) \times (1000/100)$

⇒  $x = \frac{(3 \times 342.29)}{4}$

⇒  x = 256.71 g/mol

∴ The molar mass of an unknown organic compound is 256.71 g/mol