A solution is prepared by dissolving 16.90 g of ordinary sugar (sucrose, C12H22O11, 342.3 g/mol) in 40.90 g of water. Calculate the boiling point of the solution. Sucrose is a nonvolatile nonelectrolyte.

Explanation:

The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT

ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution

Kb for water = .51°C /m

moles of sugar = 16.90 / 342.3

= .04937 moles

m = moles of sugar /  kg of water

= .04937 / .04090

= 1.207

ΔT = Kb x m

= .51 x 1.207

= .62°C .

So , boiling point of water = 100.62°C .

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.