A 20.0 % (w/w) solution of glucose (C6H12O6, molar mass = 180.16 g/mol) in water has a density of 1.0395 g/mL.

Question

A 20.0 % (w/w) solution of glucose (C6H12O6, molar mass = 180.16 g/mol) in water has a density of 1.0395 g/mL. What is its molarity and molality

Answer 1

Whenever you have a percentage question like this, it's best to assume 100 of whatever unit is concerned as the total amount. Let's say the entire solution weighs 100 grams. We then have 20.0g glucose. First, calculate the number of moles using this mass and the molecular mass of glucose.
20.0g x mol/180.16g = 0.111mol
Notice how we have grams on top and on bottom, so the units cancel to give us moles.
To calculate molarity, you take the moles of solute (substance being dissolved) and divide it by liters of solution. In this case, we're assuming 100g of solution, so we use the density of the solution that is given to us to calculate volume.
100g x mL/1.0395g = 96.2mL
Again, notice how units cancel.
Now, convert this to liters.
96.2mL x liter/1000mL = 0.0962L
Take your calculated number of moles and divide them by this number.
0.111mol/0.0962L = 1.15M

To calculate molality, you take the moles of solute and divide it by the kilograms of solvent (substance into which the solute is being dissolved; in this case, water). Since we assumed 100g of the solution, we have 80.0g of water, so we just need to convert this to kilograms.
80.0g x kg/1000g = 0.0800kg
Take your previously calculated number of moles and divide it by this number to obtain the molality value.
0.111mol/0.0800kg = 1.39m

Answer 2

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vapor pressure water at 35ºC = 0.0555atm 100ml solution x 1.08g/ml = 108g solution 5% w/w means that 5% of 108g = glucose 0.05 x 108g = 5.4g glucose 5.4g / 180g/mole = 0.03moles 108g - 5.4g = mass water = 102.6g moles water = 102.6g / 18g/mole = 5.7moles mole fraction glucose = 0.03moles / (5.7moles water + 0.03moles glucose) = 0.00523 mole fraction water = 5.7moles / 5.73moles = 0.995 Psolvent = mole fraction solvent x Pºsolvent Psolvent = 0.995 x 0.056atm = 0.0552 atm ∆T = 1.8ºC/m x m m solution = moles glucose / kg water = 0.03mole / 0.1026kg = 0.292m ∆T = 1.8ºC/m x 0.292 = 0.526ºC new temp = -0.526ºC ∆T = 0.52ºC x m solve for ∆T then add 100ºC ∏ = MRT where M = molarity, R = 0.0821L-atm/mole-K, T = Kelvin molarity = 0.03moles glucose / 0.1026L = 0.292M solve for ∏

Answer 3

W Molar Mass

Answer 4

density = 1039.5 g / L
1039.5 g/L x 20.0 / 100 =207.9 g/L
This means that there are 207.9 g of glucose pel Liter of solution
M = 207.9 g/L / 180.16 g/mol=1.15 mol/L
In 1039.5 gthere are 207.9 g of glucose => 1.15 moles
Mass water = 1039.5 - 207.9 =831.6 g => 0.8316 Kg
m = moles / Kg = 1.15 / 0.8316 =1.38

Answer 5

Let glucose be called G. Let the solution be called S

20gG/80gH2O x 1molG/180gG x 1000gH2O/1kgH2O = 1.34 mole G per 1 kg H2O = 1.34 molal (to three significant figures)

20gG/100gS x 1molG/180gG x 1.0395gS/1mLS x 1000mLS/1LS = 2.89 moles G per 1 L solution = 2.89 molar (to three sig figs)