What will be the temperature at which a solution containing 6 g of glucose per 1000 g water with boil, if molal elevation constant for water is 0.52/1000 g.
A. `1000.173^(@)`C
B. `100.0173^(@)`C
C. `100.173^(@)`C
D. None

9 views

1 Answers

Correct Answer - 2
w=6g, W=1000g, Mol. wt. of glucose= 180
`DeltaT_(b)=(1000xxK_(b)xxw)/(180xx1000)=0.0173^(@)C`
Hence boiling point of solution =b.p of water +
`DeltaT_(b)=100+0.0173^(@)C`

9 views

Related Questions