Arjun purchased 250 g of glucose (C6H12O6) for Rs 40. Find the cost of

1.	Arjun purchased 250 g of glucose (C6H12O6) for Rs 40. Find the cost of glucose per mole.
Answer» Given : Mass of urea = 250 g, Cost for 250 g glucose = Rs 40, Molecular formula of glucose = C₆H₁₂O₆ To find : Cost per mole of glucose. Calculation : Molecular formula of glucose is (C₆H₁₂O₆). Molecular mass of glucose = (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O) = (6 × 12 u) + (12 × 1 u) + (6 × 16 u) = 180 u ∴ Molar mass of glucose = 180 g mol^-1 Number of moles = \(\frac{Mass\,of\,a\,subtance}{Molar\,mass\,of\,a\,subtance}\) = \(\frac{250\,g}{180\,g\,mol^{-1}}\) = \(\frac{250}{180}\) mol Now, \(\frac{250}{180}\) mol of glucose cost = Rs 40 1 mol glucose cost = x ∴ x = \(\frac{40\times 180}{250}\) = Rs 28.8/mol of glucose ∴ The cost of glucose per mole is Rs 28.8. [Calculation using log table : \(\frac{40\times 180}{250}\) = Antilog₁₀[log₁₀(40) + log₁₀(180) + log₁₀(250)] = Antilog₁₀[1.6021 + 2.2553 – 2.3979] = Antilog₁₀[1.4595] = 28.80]

Arjun purchased 250 g of glucose (C6H12O6) for Rs 40. Find the cost of glucose per mole.

Answer»

Given :

Mass of urea = 250 g,

Cost for 250 g glucose = Rs 40,

Molecular formula of glucose = C₆H₁₂O₆

To find : Cost per mole of glucose.

Calculation : Molecular formula of glucose is (C₆H₁₂O₆).

Molecular mass of glucose

= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)

= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)

= 180 u

∴ Molar mass of glucose = 180 g mol^-1

Number of moles = \(\frac{Mass\,of\,a\,subtance}{Molar\,mass\,of\,a\,subtance}\)

= \(\frac{250\,g}{180\,g\,mol^{-1}}\)

= \(\frac{250}{180}\) mol

Now,

\(\frac{250}{180}\) mol of glucose cost = Rs 40

1 mol glucose cost = x

∴ x = \(\frac{40\times 180}{250}\)

= Rs 28.8/mol of glucose

∴ The cost of glucose per mole is Rs 28.8.

[Calculation using log table :

\(\frac{40\times 180}{250}\)

= Antilog₁₀[log₁₀(40) + log₁₀(180) + log₁₀(250)]

= Antilog₁₀[1.6021 + 2.2553 – 2.3979]

= Antilog₁₀[1.4595]

= 28.80]