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a.) Calculate the molarity of 0.150 mol of Na<sub>2</sub>S<sub> </sub>in 1.10 L of solution? b.) calculate the molarity of 32.9g of MgS in 763 mL of solution?
a) molarity = moles / volume Molarity = 0.150/1.10 = 0.136Mb) moles = 32.9/56.38 = 0.583molesVolume = 763ml = 0.763LMolarity = moles / volume Molarity = 0.583/0.763 = 0.764 M
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What is the freezing point depression of an aqueous solution of 0.75 m NaCl?
∆T = iKfmkf is molal freezing point depression constant of solvent (1.86°C/m for water)Mass of water = molar Mass of water = 18.01528g/mol × 1mol = 18.01528 g= 18.01528g H2O×1Kg...
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To make a solution, 40 g of a nonelectrolyte is dissolved into 300 g of water. The freezing point depression of the solution is measured to be -7°C. Calculate the molar mass of the solute.
46.03g/mol
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A 40.7g of a non-electrolyte is dissolved in 220g of water to make a solution. The solution's freezing point is -5.53°C. With the given data, what is the molar mass of the solute?
Molar Mass of H2O = 18.0153= 220/18.0153 = 12.212molesMolality = 12.212/-5.53= -2.21 × 40.7= 89.88g/mol
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A solution of 100.8 g of a non- dissociating solute in 135.0 g of water has a freezing point of -5.16 °C. What is the molar mass of the solute? Freezing point of water= 0 °C K-1.86 °C/m
Molar Mass of the solute=4.55M
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Calculate the freezing point and boiling point in each solution, assuming complete dissociation of the solute. a. 10.5 g FeCl3 in 1.50 * 102 g water
We Consider aqueous of FeCl3 .Depression in freezing point = i . Kf . m Elevation in Boilling point = i . Kb . m i for FeCl3 = 3...
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19.5g CH3COOH is dissolved in 500g of water the depression in the freezing point observed is 1.0k calculate van't hoff factor and dissociation constant of the acid.
ΔTfreezing=iCmK=1KCm=m( CH3COOH)/(Mr( CH3COOH)*m(H2O))=19.5/(60*0.5)=0.65 mol/kgi=1/(1.86*0.65)=0.83;Kd=(1*0.5/1.86)/(19.5/60)=0.83Answer:i=0.83;Kd=0.83
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a solution of unknown organic compound of mass 0.65gm in 27.8gm of the solvent di-phenyl gave a freezing point depression of 1.56°c
deltaT(freezing) = Cm x KfCm = n (solute) / m (solvent)To resolve this task the name of solute is required.
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Derive equation use to calculate depression in freezing point?
Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than...
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82 mg of solute is mixed into a 0.35 L solution. At a temperature of 37°C and an osmotic pressure of 3.2 x 10-3 atm, calculate the molar mass of the solute.
82 x 0.082 x 310/1000 x 3.2 x 10-3 x 0.35 = 1,861.10714
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