6. The osmotic pressure of a solution prepared by the equal volume of $8.4 \%(\mathrm{~g} / \mathrm{ml}) \mathrm{NaHCO}_{3}$ and $18 \%(\mathrm{~g} / \mathrm{ml})$ glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ is equal to the osmotic pressure of $(\quad)$. [Mr(glucose) $180\left(\mathrm{NaHCO}_{3}\right) 84$ ]
A. $5.85 \%(\mathrm{~g} / \mathrm{ml}) \mathrm{NaCl}$ solution
B. $1.5 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ sucrose solution
C. $1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ glucose solution
D. $1 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{CaCl}_{2}$ solution

Step 1 ：Since the osmotic pressure is equal, we can write the equation as $c_1M_1=c_2M_2$ where $c_1$ and $c_2$ are the concentrations of the two solutions and $M_1$ and $M_2$ are their molar masses.

Step 2 ：First, we need to find the concentration of the NaHCO_3 and glucose solution. The concentration of NaHCO_3 is $\frac{8.4}{84}=0.1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ and the concentration of glucose is $\frac{18}{180}=0.1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$

Step 3 ：Since the volumes of the two solutions are equal, the total concentration of the mixed solution is $0.1+0.1=0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1}$

Step 4 ：Now we can compare the osmotic pressure of the mixed solution with the given options. For option A, the concentration of NaCl is $\frac{5.85}{58.5}=0.1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$. For option B, the concentration of sucrose is $1.5 \mathrm{~mol} \cdot \mathrm{L}^{-1}$. For option C, the concentration of glucose is $1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$. For option D, the concentration of CaCl_2 is $1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$

Step 5 ：Comparing the osmotic pressures, we find that option D has the same osmotic pressure as the mixed solution, since $0.2 \mathrm{~mol} \cdot \mathrm{L}^{-1} = 1 \mathrm{~mol} \cdot \mathrm{L}^{-1}$

Step 6 ：Therefore, the osmotic pressure of the mixed solution is equal to the osmotic pressure of a $1 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{CaCl}_{2}$ solution. The final answer is \(\boxed{D}\)