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a sample of an unknown compound is vaporized at 200. °c. the gas produc…

Question

a sample of an unknown compound is vaporized at 200. °c. the gas produced has a volume of 1610. ml at a pressure of 1.00 atm, and it weighs 4.32 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.

Explanation:

Step1: Convert units

Convert volume from mL to L: \( V = 1610\space mL = 1.610\space L \)
Convert temperature from \(^\circ C\) to K: \( T = 200 + 273.15 = 473.15\space K \)

Step2: Use ideal gas law \( PV = nRT \) to find moles \( n \)

\( R = 0.0821\space L\cdot atm/(mol\cdot K) \), \( P = 1.00\space atm \)
\( n = \frac{PV}{RT} = \frac{1.00\space atm \times 1.610\space L}{0.0821\space L\cdot atm/(mol\cdot K) \times 473.15\space K} \)
Calculate numerator: \( 1.00 \times 1.610 = 1.610 \)
Calculate denominator: \( 0.0821 \times 473.15 \approx 38.85 \)
\( n \approx \frac{1.610}{38.85} \approx 0.04144\space mol \)

Step3: Calculate molar mass \( M \)

Molar mass \( M = \frac{mass}{moles} = \frac{4.32\space g}{0.04144\space mol} \approx 104\space g/mol \) (rounded to 3 significant digits)

Answer:

\( 104\space g/mol \)