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Chemistry

Answered

How many grams of glucose (C6H12O6) are contained in 555 mL of a 1.77 M glucose solution?

Sagot :

Goldenid Goldenid

molarity =wt/M.wt * 1000/V

=> 1.77 = wt/180 * 1000/555

> wt = 1.77 * 180*555 /1000 = 176.823 g

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