High School

How many grams of glucose, dissolved in 350.0 g of water, are needed to make a 1.55 m solution?

A. 97.7
B. 45.5
C. 279
D. 145

Question 4: What is the freezing point in ∘C of a 0.45 m solution of glucose, \[C_6H_{12}O_6\], in water? (The \[K_f\] for water is 1.858 ∘C/m.)

A. -0.42
B. 0.84
C. -1.24
D. -0.84

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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