Answer :
279.72 grams of glucose is needed to make a 1.55 m solution in 350g of water. The freezing point of the solution is -0.84 °C.
Molarity (m) = moles of solute/liters of solution
The moles of glucose required:
Given molarity (m) = 1.55 m
The volume of water ([tex]V_w[/tex]) = 350.0 g
(We assume water density is 1 g/mL, so 350.0 g ≈ 350.0 mL)
Molar mass of glucose = 180.16 g/mol
The moles of glucose is given by;
moles of glucose = molarity * liters of solution
The volume of solution = Volume of water + Volume of glucose
= 350.0 mL + Volume of glucose ( 1.55 m = 1.55 mol/L)
Volume of glucose = 1.55 mol / 1 L
= 1.55 L
moles of glucose = 1.55 mol
molar mass of glucose = 180.16 g/mol
grams of glucose = 1.55 mol * 180.16 g/mol
= 279.72 g
4) ΔT = Kf * m
ΔT = freezing point depression (in °C)
Kf = freezing point depression constant for water (1.858 °C/m)
m = molality of the solution (in mol/kg)
m = 0.45 m,
ΔT = 1.858 °C/m * 0.45 m
= 0.8361 °C
Freezing point = 0 °C - 0.8361 °C
= -0.84 °C
Therefore, the freezing point of the solution is approximately -0.84 °C, and 279.72 grams of glucose is needed to make a solution in water.
Learn more about freezing point from the given link.
https://brainly.com/question/31357864
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