Answer :
Final answer:
To prepare 235.0 mL of 22%w/v glucose (C6H12O6), the mass of glucose needed is 51.7 grams.
So, option a) 51.7 grams is the correct answer.
Explanation:
To calculate the mass of glucose needed, we need to first determine the number of moles of glucose in the solution. The solution is 22% w/v, which means there are 22 grams of glucose in 100 mL of solution. To convert mL to L, divide by 1000. Therefore, in 235.0 mL of solution, there will be (22/100) * 235.0 = 51.7 grams of glucose needed. So, option a) 51.7 grams is the correct answer.
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