11. There are four water solutions of the equal volume in which there are equal mass of glucose, $\mathrm{CaCl}_{2}, \mathrm{NaHCO}_{3}$ and sucrose respectively. Then whose freezing-point is the lowest? ( ) $\left[\mathrm{Mr}(\right.$ glucose $\left.) 180(\mathrm{HAc}) 60\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right) 106\left(\mathrm{CaCl}_{2}\right) 111\right]$ A. Glucose B. HAc C. $\mathrm{CaCl}_{2}$ D. $\mathrm{Na}_{2} \mathrm{CO}_{3}$

Step 1 ：Calculate the moles of each solute: \(moles = \frac{mass}{molar\ mass}\)

Step 2 ：moles_glucose = \(\frac{180}{180} = 1\)

Step 3 ：moles_HAc = \(\frac{180}{60} = 3\)

Step 4 ：moles_Na2CO3 = \(\frac{180}{106} \approx 1.70\)

Step 5 ：moles_CaCl2 = \(\frac{180}{111} \approx 1.62\)

Step 6 ：Calculate the number of particles for each solute:

Step 7 ：particles_glucose = 1

Step 8 ：particles_HAc = 3

Step 9 ：particles_Na2CO3 = 1.70 * 2 = 3.40

Step 10 ：particles_CaCl2 = 1.62 * 3 = 4.86

Step 11 ：Compare the number of particles and find the lowest freezing point: \(\boxed{\text{A. Glucose}}\)