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Calculate the number of moles and molecules of urea present in 5.6 g of urea. |
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Answer» Given : Mass of urea = 5.6 g To find : The number of moles and molecules of urea Formulae : i. Number of moles = \(\frac{Mass\,of\,a\,substance}{Molar\,mass\,of\,a\,substance}\) ii. Number of molecules = Number of moles × Avogadro’s constant Mass of urea = 5.6 g Molecular mass of urea, NH2CONH2 = (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O) = (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) = 60 u ∴ Molar mass of urea = 60 g mol-1 ∴ Number of moles = \(\frac{Mass\,of\,a\,substance}{Molar\,mass\,of\,a\,substance}\) = \(\frac{5.6g}{60g\,mol^{-1}}\) = 0.09333 mol [Calculation using log table : \(\frac{5.6}{60}\) = Antilog10[log10(5.6) – log10(60)] = Antilog10 [0.7482 – 1.7782] = Antilog10 \(\overline{2} .9700\) = 0.09333] Now, Number of molecules of urea = Number of moles × Avogadro’s constant = 0.09333 mol × 6.022 × 1023 molecules/mol = 0.5616 × 1023 molecules (by using log table) = 5.616 × 1022 molecules ∴ Number of moles of urea = 0.0933 mol Number of molecules of urea = 5.616 × 1022 molecules [Calculation using log table : 0.09333 × 6.022 = Antilog10[log10(0.09333) + log10(6.022)] = Antilog10[ \(\overline{2} .9698\) + 0.7797] = Antilog10 \(\overline{1} .7495\) = 0.5616] |
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