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What is the molality of a solution which contains 36 g of glucose `(C_(4)H_(12)O_(5))` in 250 g of water ?

Answer» Mass of glucose = 36 g
Molar mass of glucose `= 6 xx 12+12xx1+6xx16=180 "amu"=180 " g mol"^(-1)`
Mass of water `= 250 g = (250)/(1000)=0.25 kg`
Molarity of solution `(m)=("Mass of glucose/Molar mass")/("Mass of water in kg")`
`= ((36g))/((180" g mol"^(-1))xx(0.25 kg))=0.8"mol kg"^(-1)=0.8`m.


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