Question

A 5% solution (w/w) of cane sugar (molar mass=342g) has freezing point 271K. What will be the freezing point of 5% glucose (molar mass= 180g) in water if the freezing point of pure water is 273.15K?
(A) 273.07K
(B) 268.07K
(C) 273.15K
(D) 260.09K

Answer 1

Hint : We will calculate the molal freezing constant (${{K}_{f}}$) for water using the relation ${{K}_{f}}=\dfrac{\Delta {{T}_{f}}}{m}$ and then we will calculate the freezing point of glucose using, ${{T}_{glu\cos e}}={{T}_{water}}-{{K}_{f}}\times m$.
${{T}_{water}}=273.15K$

Complete step by step solution :
Let’s take case 1 and calculate the molal freezing constant (${{K}_{f}}$) for water using cane sugar
The formula for calculating ${{K}_{f}}$
${{K}_{f}}=\dfrac{\Delta {{T}_{f}}}{m}$…….equation 1
Now, $\Delta {{T}_{f}}={{T}_{water}}-{{T}_{cane sugar}}$
$\therefore ,\Delta {{T}_{f}}=273.15-271=2.15$…..equation 2
Now, we will calculate the molality of cane sugar.
A 5% solution of cane sugar means 5g cane sugar is dissolved in 95g of water.
Therefore, given mass=5g
Molar mass=342g
Wt of solvent=95g

Formula for molality is:

& w{{t}_{solvent}} \\\ & \\\ \end{aligned}}$$ $$m=\dfrac{{{m}_{solute}}\times 1000}{\begin{aligned} & w{{t}_{solvent}} \\\ & \\\ \end{aligned}}=\dfrac{5\times 1000}{342\times 95}$$ $$m=0.153$$…..equation 3 Where, ${{m}_{solute}}=\dfrac{Mas{{s}_{given}}}{MolarMass}=\dfrac{5}{342}$ Now, we will put the value of equation 2 and equation 3 in equation 1 and calculate ${{K}_{f}}$ Therefore, ${{K}_{f}}=\dfrac{2.15}{0.153}=14.05$…..equation 4 Now, let’s take case 2 and calculate the freezing point of 5% glucose Now, we will calculate the molality of glucose. A 5% solution of cane sugar means 5g glucose is dissolved in 95g of water. Therefore given mass=5g Molar mass=180g Wt of solvent=95g Formula for molality is: $$m=\dfrac{{{m}_{solute}}\times 1000}{\begin{aligned} & w{{t}_{solvent}} \\\ & \\\ \end{aligned}}$$ $$m=\dfrac{{{m}_{solute}}\times 1000}{\begin{aligned} & w{{t}_{solvent}} \\\ & \\\ \end{aligned}}=\dfrac{5\times 1000}{180\times 95}$$……where, $${{m}_{solute}}=\dfrac{Mas{{s}_{given}}}{MolarMass}=\dfrac{5}{180}$$ $$m=0.2923$$…..equation 5 Now, we will calculate the freezing point of glucose using the formula $$\Delta {{T}_{f}}={{K}_{f}}\times m$$ $$\Delta {{T}_{f}}={{T}_{water}}-{{T}_{glu\cos e}}$$ Therefore, ${{T}_{glu\cos e}}={{T}_{water}}-{{K}_{f}}\times m$ …..equation 6 On putting the values of equation 4 and equation 5 in equation 6, we get $${{T}_{glu\cos e}}=\,273.15-14.05\times 0.2923$$ Therefore, ${{T}_{glu\cos e}}=269.04K$, Which is close to option (B). **So, the correct answer is “Option B”.** **Note** : Convert the temperatures into kelvin. Take weight of solution in Kilograms. Freezing point of a solute is always less than the freezing point of its solvent.