Question

`18 g` of glucose `(C_(6)H_(12)O_(6))` is dissolved in `1 kg` of water in a saucepan. At what temperature will the water boil (at 1 atm) ? `K_(b)` for water is `0.52 K kg mol^(-1)`.

Answer 1

`W_1=" Weight of solvent "(H_2O)=1kg`.
`W_2= " Weight of solute "(C_6H_12O_6)=18 gm`
`M_2=" Molar mass of solute "(C_6H_12O_6)=180 g//mol`
`K_b=0.52K kg//mol`
`T_b^@=373.15 K`
`DeltaT_b=(K_bxx1000xxW_2)/(M_2xxW_1)=(0.52xx1000xx18)/(180xx1000)=0.052 K`.
`Delta T_b=T_b-T_b^@`
`0.052=T_b-373.15`
`T_b=373.202 K`.