glucose is a sugar chemical formula C6H12O6. calculate
…
i) concentration of glucose in ppm if the solution was prepared by dissolving 163g to produce 750ml.
ii) molarity glucose solution above(i)
iii) mass of glucose in 25ml solution if concentration of the solution is 5.6% (w/v)
The Correct Answer and Explanation is :
Calculated Results:
- Concentration of glucose in ppm: ( 2.173 \times 10^8 \, \text{ppm} )
- Molarity of the glucose solution: ( 1.21 \, \text{M} )
- Mass of glucose in 25 mL solution: ( 1.4 \, \text{g} )
Explanation:
- Concentration in ppm:
Parts per million (ppm) is calculated as:
[
\text{ppm} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (L)}} \right) \times 10^6
]
Here, 163 g of glucose is dissolved in 750 mL of solution (( 0.75 \, \text{L} )). Substituting:
[
\text{ppm} = \left( \frac{163}{0.75} \right) \times 10^6 = 2.173 \times 10^8 \, \text{ppm}
] - Molarity of glucose solution:
Molarity is defined as:
[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution (L)}}
]
To find moles of glucose, we use its molar mass:
[
\text{Molar mass of glucose} = 6(12) + 12(1) + 6(16) = 180 \, \text{g/mol}
]
Moles of glucose:
[
\text{Moles} = \frac{163}{180} \approx 0.906 \, \text{mol}
]
Substituting in the molarity formula:
[
\text{Molarity} = \frac{0.906}{0.75} \approx 1.21 \, \text{M}
] - Mass of glucose in 25 mL solution:
For a 5.6% (w/v) solution, 5.6 g of glucose is present in 100 mL of solution. For 25 mL:
[
\text{Mass} = \frac{5.6}{100} \times 25 = 1.4 \, \text{g}
]
These calculations provide insights into solution concentration metrics in ppm, molarity, and weight/volume percentages.