A glucose solution is prepared by dissolving 5.10 g of glucose, C6H12O6, in 110.5 g of water. What is the molality of the glucose solution? A) 0.283 m B) 0.000256 m C) 0.245 m D) 0.256 m

Question :- A glucose solution is prepared by dissolving 5.10 g of glucose, C6H12O6, in 110.5 g of `water. What is the molality of the glucose solution?
A) 0.283 m B) 0.000256 m C) 0.245 m D) 0.256 m

Answer :- 

 molality = number of moles of solute /mass of solvent in Kg 

The molar mass of glucose is   = 6*C + H*12+ O*6: 

                                                     =180.16 g/mol

So number of moles  of glucose = 5.10 / 180 = 0.0283 mol

These are present in 5.10g + 110.5g of solution = 115.6g

divide by 1000 to convert in Kg 

so 115.6 gram = 0.1156 Kg
 

molality = 0.0283 moles / 0.1156 kg
 

molality = 0.256                                            Answer  = D