What is the mass percent of sucrose (C12H22O11, Mm=342 g/mol) in a 0.371-m sucrose solution?

Answer 1

.372m=mass/342)/kg of solvent
kg of solvent= 1-kgofsoute=(in grams(1000-mass)
.372*342=mass/1-mass/1000)=1000*mass/(1000-mass)
127*(1000-mass)=1000mass
127000=1000mass+127mass
mass= 127000/1127 grams or 113 grams.
check my math.