A sample of an unknown compound is vaporized at 150*C. The gas produced has a volume of 1240mL at a pressure of 1.00atm, and it weighs 3.72g. Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Round your answer to 3 significant digits. **Understanding Molar Mass Calculations Using the Ideal Gas Law**A sample of an unknown compound is vaporized at 150°C. The gas produced has a volume of 1240 mL at a pressure of 1.00 atm, and it weighs 3.72 g.Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Round your answer to 3 significant digits.---**Details of the Assessment:**- **Volume of Gas:** 1240 mL- **Pressure:** 1.00 atm- **Mass of Gas:** 3.72 g- **Temperature:** 150°C**Steps to Calculate:**1. **Convert the Volume from mL to L:** \( \text{Volume in L} = \text{Volume in mL} \times \frac{1}{1000} \)2. **Convert Temperature to Kelvin:** \( \text{Temperature in K} = \text{Temperature in °C} + 273.15 \)3. **Use the Ideal Gas Law:** \( PV = nRT \) - \( P \) = Pressure in atm - \( V \) = Volume in L - \( n \) = Number of moles - \( R \) = Ideal gas constant \( (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}) \) - \( T \) = Temperature in K4. **Calculate the Number of Moles (\( n \)):** Rearrange the ideal gas law to solve for \( n \): \[ n = \frac{PV}{RT} \]5. **Calculate the Molar Mass (\( M \)):** \[ M = \frac{\text{mass of gas}}{n} \]**Graphical Interface:**The interface includes input fields for the gas measurement and a "Check" button to verify calculations. **Resources and Help:**- **Explanation and Guidance:** Additional explanations are available for each step.- **Checking Facility:** Use the "Check" button to validate your answer.Ensure precision in your calculations and round off your final result to three significant digits as needed.

Given temperature (T) = 150°C =150°C + 273.15 = 423.15K Given volume occupied by the gas (V) = 1240 ml = 1.240 L Given pressure (P) = 1.0 atm Given mass of the compound (m) = 3.72 g Number of moles of the compound (n) = ? Universal gas constant (R) = 0.0821 L.atm.K-1.mol-1 According to the ideal gas equation, PV = nRTn = PVRTn = 1.0 atm×1.240 L0.0821 L·atm·K-1·mol-1×423.15 Kn = 0.036 mol So, the number of moles of the given gas (n) = 0.036 molNumber of moles = given massmolar mass0.036 mol = 3.72 gmolar massMolar mass = 3.72 g0.036 mol=103.33 g/mol≃1.03×102 g/mol …………(rounded off to 3 significant digits) Hence, the molar mass of the given compound is ≃ 1.03×102 g/mol.