Solution: How many moles of CO(NH 2)2 are present in 131 g of water, if the freezing point of the solution is −4.02 °C? kf (water) =1.86 C • kg/mol.1. 0.263 mol2. None of these3. 0.280 mol4. 0.297 mol5. 0.288 m

Solution: How many moles of CO(NH 2)2 are present in 131 g of water, if the freezing point of the solution is −4.02 °C? kf (water) =1.86 C • kg/mol.1. 0.263 mol2. None of these3. 0.280 mol4. 0.297 mol5. 0.288 m

How many moles of CO(NH _{2})_{2} are present in 131 g of water, if the freezing point of the solution is −4.02 °C? k_{f} (water) =1.86 C • kg/mol.

1. 0.263 mol

2. None of these

3. 0.280 mol

4. 0.297 mol

5. 0.288 mol

6. 0.272 mol

We’re being asked to **determine the moles of CO(NH _{2})_{2}** present in

Recall that the freezing point of a solution is *lower* than that of the pure solvent and the ** change in freezing point (ΔT_{f})** is given by:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{pure}\mathbf{}\mathbf{solvent}}{\mathbf{-}}{{\mathbf{T}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{solution}}}$

The ** change in freezing point** is also related to the molality of the solution:

$\overline{){{\mathbf{\Delta T}}}_{{\mathbf{f}}}{\mathbf{=}}{{\mathbf{imK}}}_{{\mathbf{f}}}}$

where:

**i** = van’t Hoff factor

**m** = molality of the solution (in m or mol/kg)

**K _{f}** = freezing point depression constant (in ˚C/m)

Recall that the ** molality of a solution** is given by:

$\overline{){\mathbf{molality}}{\mathbf{=}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{kg}\mathbf{}\mathbf{solvent}}}$

**For this problem, we need to do the following:**

**Step 1:**** **Calculate for ΔT_{f}.

* Step 2:* Determine the molality of the solution.

* Step 3:* Calculate the mass of CO(NH