Weekly Solution Calculations 11. To a 10.0 ml fermentation culture tube, you add

Question

Weekly Solution Calculations 11. To a 10.0 ml fermentation culture tube, you add 1.0 ml of 10% glucose MW 180, 0.5 ml of sucrose (MW 342.30), and 8.5 ml of an aqueous yeast culture. What is the molarity of glucose in the tube? 12. If0.030 moles of CO, are produced in a yeast culture fermentation supplied with glucose (MW180), how many grams of glucose were consumed? 13. If given a 2.0 M maltose stock solution, how much stock and how much water are needed to make 500 ml of a 10% solution? (MW 342.30)

Explanation / Answer

1. A 10% (w/v) glucose solution would be 10 g of glucose in 100ml of water. If there is 10g of glucose in 100ml of water, then there will be 100g in 1 litre of water (molarity is just moles per liter.)

The molecular weight of glucose is: 180g/mol

The molarity is therefore: 100 / 180 = 0.555 M of glucose

2. For example if this this yeast experiments produce 500 ml (0.5 liters) of gas (CO2) in the glucose fermentation. The volume produced is equivalent to (0.5 liters) • (0.030 mole/liter of glucose) = 0.015 moles of CO2 produced. So if 0.015 moles of CO2 are produced in our fermentation example, then 0.015 moles of glucose are produced.

3. 2M maltose solution there.

To prepare 10%solution, 10g of maltose in 100ml and 100g of maltose in 1 litre of water.

1M= 342.30g, therefore 100g is 0.29M in 1litre of water

Now we have

x X 2M= 500mlX 0.29M (which is 10% solution)

x= 500x2.9/2 = 73ml

Therefore 73ml of 2M maltose is made upto 500ml by adding 423ml of water.

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.