18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? (Kb for water = 0.52 K kg mol-1)

Question

18 g of glucose, C6H12O6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? (Kb for water = 0.52 K kg mol-1) (A) 273.202 K (B) 173 K (C) 373.202 K (D) 300 K

Tardigrade Admin · Accepted Answer

Δ Tb = (Kb × 1000 × w2/M2 × w1) = (0.52 × 1000 × 18/180 × 1000) = 0.052 Boiling point of solution T = T0 + 0.052 = 373.15 + 0.052 = 373 .202 K

Q. $18 g$ of glucose, $C_{6} H_{12} O_{6}$ , is dissolved in $1 k g$ of water in a saucepan. At what temperature will water boil at $1.013$ bar? ( $K_{b}$ for water $= 0.52 K k g m o l^{- 1}$ )

$273.202 K$

$173 K$

$373.202 K$

$300 K$

$Δ T_{b} = \frac{K _{b} \times 1000 \times w _{2}}{M _{2} \times w _{1}}$
$= \frac{0.52 \times 1000 \times 18}{180 \times 1000} = 0.052$
Boiling point of solution $T = T_{0} + 0.052$
$= 373.15 + 0.052 = 373.202 K$

Q. 18g of glucose, C6​H12​O6​, is dissolved in 1kg of water in a saucepan. At what temperature will water boil at 1.013 bar? (Kb​ for water =0.52Kkgmol−1)

273.202K

173K

373.202K

300K