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? ( ) $[\mathrm{Mr}($ glucose $)180(\mathrm{HAc})60\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right)106\left({\mathrm{CaCl}}_2\right)111]$
A. Glucose
B. HAc
C. ${\mathrm{CaCl}}_2$
D. ${\mathrm{Na}}_2{\mathrm{CO}}_3$

Step 1 ：Calculate the moles of each solute: $moles=\frac{mass}{molarmass}$

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: $\boxed{{\displaystyle \text{A. Glucose}}}$