Question

1 mole glucose is added to 1 L of wat
(1) 373.512°C
(3) 99.488°C
bl2°) = 0.512 K kg mole-1, boiling point of solution will be
(2) 100.512°C
(4) 372.488°C
value of. constant de​

Answer 1

Answer:

weight on 1 L of water is 1kg

moles of glucose 1moles

elevation in boling point =(kb)(moles of glucose)/(weight of water in kg)=.512×1/1=.512 k theirfore new boling point is 373.512k or 100.512(I for glucose is 1 because it neither dissociate or ionizes in water and it is also monosaccharide and elevation in boiling point in kelvin and in Celsius will be same)

Answer 2

Explanation:

l think it is correct answer and helpful