Answer:3.5 ml of water will be required to sterile the 700 mg of Kanamycin.
Step-by-step explanation:
Mass of te Kamamycin = 700 mg = 0.7 g
1000 mg = 1 g
1 grams of Kanamycin is sterile by 5 mL of water as stated in label
Then ,0.7 g of water will be sterile by =
![0.7 g* 5ml=3.5 mL](https://img.qammunity.org/2019/formulas/chemistry/college/np4o0gvurq9ce56pitq3c5ptca2sffq2xu.png)
3.5 ml of water will be required to sterile the 700 mg of Kanamycin.