Studyledge

States of Matter: Question 2

Syllabus 4.1

Structured AS 8 marks

A technician is finding the molar mass of a volatile organic liquid, X, using a heated gas syringe. A few drops of liquid X, of mass 0.145 g, are injected into the syringe and allowed to vaporise completely. The resulting gas occupies 42.5 cm³, measured at a pressure of 101 kPa and a temperature of 100 °C.

(a) Convert the volume, pressure and temperature of the gas produced into the SI units required to use the ideal gas equation. [3]

(b) Use the ideal gas equation pV=nRTpV = nRT, where R=8.31 J K−1mol−1R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}, to calculate the amount, in mol, of gas produced. [3]

(c) Hence calculate the molar mass of X, giving your answer to 3 significant figures. [2]

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Worked solution

Part (a): Converting to SI units

The ideal gas equation requires pressure in pascals (Pa), volume in cubic metres (m3\text{m}^3) and temperature in kelvin (K).

Volume: since 1 m3=106 cm31\ \text{m}^3 = 10^6\ \text{cm}^3, V=42.5 cm3=42.5106 m3=4.25×10−5 m3V = 42.5\ \text{cm}^3 = \frac{42.5}{10^6}\ \text{m}^3 = 4.25\times10^{-5}\ \text{m}^3

Pressure: since 1 kPa=103 Pa1\ \text{kPa} = 10^3\ \text{Pa}, p=101 kPa=101×103 Pa=1.01×105 Pap = 101\ \text{kPa} = 101\times10^{3}\ \text{Pa} = 1.01\times10^{5}\ \text{Pa}

Temperature: adding 273 to convert from Celsius to kelvin, T=100+273=373 KT = 100 + 273 = 373\ \text{K}

Part (b): Calculating the amount, in mol, of gas

Rearranging pV=nRTpV = nRT for nn: n=pVRTn = \frac{pV}{RT}

Substituting the SI values from part (a) and R=8.31 J K−1mol−1R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}: n=(1.01×105)×(4.25×10−5)8.31×373n = \frac{(1.01\times10^{5})\times(4.25\times10^{-5})}{8.31\times373}

Working out the numerator and denominator separately: pV=1.01×105×4.25×10−5=4.2925pV = 1.01\times10^{5}\times4.25\times10^{-5} = 4.2925 RT=8.31×373=3099.63RT = 8.31\times373 = 3099.63

So: n=4.29253099.63=1.3848×10−3 moln = \frac{4.2925}{3099.63} = 1.3848\times10^{-3}\ \text{mol}

To 3 significant figures, n=1.38×10−3 moln = 1.38\times10^{-3}\ \text{mol}.

(Check: repeating the division independently, 42925÷3099.63≈13.8542925 \div 3099.63 \approx 13.85, so n≈13.85×10−4=1.385×10−3 moln \approx 13.85\times10^{-4} = 1.385\times10^{-3}\ \text{mol}. Consistent with the value above.)

Part (c): Calculating the molar mass

The amount of substance is related to mass and molar mass by n=mMrn = \dfrac{m}{M_r}, so: Mr=mn=0.1451.3848×10−3M_r = \frac{m}{n} = \frac{0.145}{1.3848\times10^{-3}}

Mr=104.7 g mol−1M_r = 104.7\ \text{g mol}^{-1}

To 3 significant figures: Mr=105 g mol−1M_r = 105\ \text{g mol}^{-1}

Final answers

  • (a) V=4.25×10−5 m3V = 4.25\times10^{-5}\ \text{m}^3, p=1.01×105 Pap = 1.01\times10^{5}\ \text{Pa}, T=373 KT = 373\ \text{K}
  • (b) n=1.38×10−3 moln = 1.38\times10^{-3}\ \text{mol}
  • (c) Mr=105 g mol−1M_r = 105\ \text{g mol}^{-1} (to 3 s.f.)