SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & mass_{\text{solute}} = \text{2.47 g} \\ & V_{\text{solution}} = \text{202 mL = 0.202 L} \\ & \pi = \text{8.63 mmHg} \times \frac{\text{1 atm}}{\text{760 mmHg}} = \text{0.011355 atm} \\ & T = 21^{\circ}\text{C} + 273.15 = \text{294.15 K} \end{aligned}[/tex]Step 2: Calculate the molarity of the solution.[tex]\begin{aligned} M & = \frac{\pi}{RT} \\ & = \frac{\text{0.011355 atm}}{\left(0.082057 \: \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(\text{294.15 K})} \\ & = 4.7044 \times 10^{-4} \: \text{mol/L} \end{aligned}[/tex]Step 3: Calculate the number of moles of solute (polymer).[tex]\begin{aligned} n_{\text{solute}} & = (M)(V_{\text{solution}}) \\ & = (4.7044 \times 10^{-4} \: \text{mol/L})(\text{0.202 L}) \\ & = 9.5029 \times 10^{-5} \: \text{mol} \end{aligned}[/tex]Step 4: Calculate the molar mass of solute.[tex]\begin{aligned} MM_{\text{solute}} & = \frac{mass_{\text{solute}}}{n_{\text{solute}}} \\ & = \frac{\text{2.47 g}}{9.5029 \times 10^{-5} \: \text{mol}} \\ & = \boxed{2.60 \times 10^4 \: \text{g/mol}} \end{aligned}[/tex]Hence, the molar mass of the polymer is 2.60 × 10⁴ g/mol.
Answer:III - Underline the informal words and change them to formal words:1. He's gonna be angry. → He is going to be angry.2. I wanna learn how to ski. → I want to learn how to ski.3. Didja like the movie? → Did you like the movie?4. We hafta leave now. → We have to leave now.5. I bought ya apples and grapes. → I bought you apples and grapes.IV - Simple sentences using Simile:1. Her smile was as bright as the sun.2. He ran like a cheetah during the race.