Question

Calculate the number of moles and molecules of urea present in 5.6 g of urea.

Answer 1

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]