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What is the molarity of a 10.2 % by mass glucose (c6h12o6) solution? (the density of the solution is 1.03 g/ml .)?

Respuesta :

W0lf93
Molality = moles of solute divided by kilograms of solvent In this case let’s assume a 100g sample solution: Since its 10.5% by mass, there is 10.5g glucose. The rest of the 89.5g is water. Divide the 10.5g glucose by its molar mass (180g) to get the # of moles, which is 0.058moles So now we have 0.058mole glucose divided by 0.089kilograms of water: the answer is 0.65molal

Answer:

0.57 m

Explanation:

What is the molarity of a 10.2 % by mass glucose (c6h12o6) solution? (the density of the solution is 1.03 g/ml .)?

Data:

m=?

M= 0.057

V= 100 gr.

Glucose/Molar Mass: 180.156 g/mol

Formula:

(Percentage  by  mass  written as ratio)

(from molar mass of glucose)

[tex]\frac{(1 mL)}{(1.03 gr)}[/tex](from given density of the solution)

Procedures:

1. Start with the numerator 10.20 gr (C6H12O6) and convert the molar mass to moles (C6H12O6):

Substance formula: (C6H12O6)

Mass of the substance in grams: 10.20

Amount of the substance, in moles: 0.057

Molar mass of the substance: 180.156

Molar mass information:

72.066 C (6*12.011) + 12.096 H (12*1.008) + 95.994 O (6*15.999)

2. Continue with the denominator (100 g of solution) and convert them to mL of solution (Using density), then convert them to liters.

100 g = 100 ml

density of water: 1 g/mL

100 ml = 0.1 L

3. Finally, divide the moles (C6H12O6) by Liters of the solution to get the molarity:

[tex]\frac{0.057}{0.1}[/tex]=0.57 m

Source: Page 53/100 and 54/100

Nivaldo J. Tro. (2014). Converting between Concentration Units. 15 de noviembre del 2021, de Pearson Education, Inc Sitio web: https://chem.ntou.edu.tw/ezfiles/28/1028/attach/22/pta_6182_1648896_43084.pdf

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