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istory Book marks People Window Help a https://session.masteringchemistry.com/myct/itemView/offset-next&assignmentProblemID-92928137; MasteringChermistry CH12 HW CH12HW Practice Problem 12.15 ssio Practice Problem 12.15 Nora Part A How many milliters of a 2.67 M solution of glucose, Ce H120s. molar mass 180.155 g/ Express your answer with the appropriate units. mol, do you nood to obtain 250 g of glucose edit for v Value ml Comp Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

given, Molarity = 2.67 M
let volume required be v in L
so,
number of mole of glucose = (molarity of glucose)*(volume of solution in L)
= 2.67*v

also,
molar mass of glucose = 180 g/mol
number of mole of glucose = (given mass)/(molar mass of glucose)
so,
2.67*v = 25.0/180
v = 0.0520 L
= 52.0 mL

Answer : 52.0 mL

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