What is the freezing point of a solution that contains text10 text.0 textg of a glucose ( C 6 H 12 O 6) in 100 g of H 2 O ? K f =1.86° C / m

Question

What is the freezing point of a solution that contains text10 text.0 textg of a glucose ( C 6 H 12 O 6) in 100 g of H 2 O ? K f =1.86° C / m (A) -0.186° C (B) +0.186° C (C) -0.10° C (D) -1.03° C

Tardigrade Admin · Accepted Answer

Δ Tf=Kfm=Kf× (WB/MB . WA) =(1.86 text K kg mo ( textl)- 1 × 10.0 text g/180 text g mo ( textl)- 1 × 100 text g × ((1 text kg)1000 text g)) =(1.86 × 10.0 × 1000 text K/180 × 100) =1.03 text K=1.03° C F.P. of solution =F.P of water -Δ Tf =0° C-1.03° C=-1.03° C

Q. What is the freezing point of a solution that contains $10 .0 g$ of a glucose $(C_{6} H_{12} O_{6})$ in $100 g$ of $H_{2} O ? K_{f} = 1.8 6^{\circ} C / m$

$- 0.18 6^{\circ} C$

$+ 0.18 6^{\circ} C$

$- 0.1 0^{\circ} C$

$- 1.0 3^{\circ} C$

Q. What is the freezing point of a solution that contains 10.0g of a glucose (C6​H12​O6​) in 100g of H2​O?Kf​=1.86∘C/m

−0.186∘C

+0.186∘C

−0.10∘C

−1.03∘C

ΔTf​=Kf​m=Kf​×MB​.WA​WB​​ =180 g mo(l)−1×100 g×(1000 g1 kg​)1.86 K kg mo(l)−1×10.0 g​ =180×1001.86×10.0×1000 K​ =1.03 K=1.03∘C F.P. of solution =F.P of water −ΔTf​ =0∘C−1.03∘C=−1.03∘C

Q. What is the freezing point of a solution that contains $10 .0 g$ of a glucose $(C_{6} H_{12} O_{6})$ in $100 g$ of $H_{2} O ? K_{f} = 1.8 6^{\circ} C / m$

$- 0.18 6^{\circ} C$

$+ 0.18 6^{\circ} C$

$- 0.1 0^{\circ} C$

$- 1.0 3^{\circ} C$