Question

a solution of urea was found to be isotonic with a solution of salt XY of molecular weight 74.6. if .15 moles of urea are dissolved in a certain volume V ml of the isotonic solution , the amount (in gm) of salt in the solution will be :

Answer 1

5.595

Answer 2

When two solutions are isotonic, their osmotic pressures are equal. The osmotic pressure ($\Pi$) is given by the formula $\Pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature.

For isotonic solutions at the same temperature, we have: $i_1 C_1 = i_2 C_2$

Let's identify the solutes and their properties:

1. Urea ($CO(NH_2)_2$): Urea is a non-electrolyte, meaning it does not dissociate into ions in solution. Therefore, its van't Hoff factor ($i_{\text{urea}}$) is 1.
2. Salt XY: The molecular weight of salt XY is given as 74.6 g/mol. Since it's a salt, it is expected to be an electrolyte and dissociate into ions. Assuming it's a 1:1 electrolyte (e.g., $X^+Y^-$), it dissociates into two ions ($X^+$ and $Y^-$). Therefore, its van't Hoff factor ($i_{\text{XY}}$) is 2 (assuming complete dissociation). A common salt with molecular weight 74.6 g/mol is KCl (K=39.1, Cl=35.45, sum=74.55), which indeed dissociates into two ions ($K^+$ and $Cl^-$).

The problem states that 0.15 moles of urea are dissolved in a certain volume V ml, and this solution is isotonic with a solution of salt XY (implying the salt XY solution is also in the same volume V ml).

Let $C_{\text{urea}}$ be the molar concentration of the urea solution and $C_{\text{XY}}$ be the molar concentration of the salt XY solution. $C_{\text{urea}} = \frac{\text{moles of urea}}{\text{Volume (in L)}} = \frac{0.15 \text{ moles}}{V \text{ ml}} = \frac{0.15}{V/1000} \text{ M}$ Let $n_{\text{XY}}$ be the moles of salt XY in the solution. $C_{\text{XY}} = \frac{\text{moles of XY}}{\text{Volume (in L)}} = \frac{n_{\text{XY}} \text{ moles}}{V \text{ ml}} = \frac{n_{\text{XY}}}{V/1000} \text{ M}$

Now, apply the isotonic condition: $i_{\text{urea}} C_{\text{urea}} = i_{\text{XY}} C_{\text{XY}}$ $1 \times \left(\frac{0.15}{V/1000}\right) = 2 \times \left(\frac{n_{\text{XY}}}{V/1000}\right)$

The volume term ($V/1000$) cancels out from both sides: $0.15 = 2 \times n_{\text{XY}}$ $n_{\text{XY}} = \frac{0.15}{2} = 0.075 \text{ moles}$

To find the amount (in gm) of salt XY, multiply the moles by its molecular weight: Amount of salt XY = $n_{\text{XY}} \times \text{Molecular weight of XY}$ Amount of salt XY = $0.075 \text{ moles} \times 74.6 \text{ g/mol}$ Amount of salt XY = $5.595 \text{ g}$