The same mass of solute is dissolved in 15.0 grams of cyclohexane instead of t-butanol.?
A) Use the freezing point depression equation ∆Tf = Kf*m. Kf is just a constant, there's one for most solvents. Note that m is the MOLALITY of the solution, not molarity. First, find the change in temperature between the freezing point of pure t-butanol and the freezing point of the solution. ∆Tf = (freezing pt. of t-butanol) - (freezing pt. of solution) ∆Tf = 25.50 °C - 23.9 °C = 1.60 °C. Now you can find the molality of the solution, given that the Kf of t-butanol is 9.1 °C/m (I just googled it). ∆Tf = Kf*m ---> m = ∆Tf / Kf = 1.60 °C / 9.1 = 0.176 m. B) molar mass is g/mol by definition. We know the grams of our solute, it's .442 g. We just need to find the moles of the solute. You calculate the moles of solute from the molality of the solid that we calculated earlier. molality is defined as (moles of solute) / (kg of solvent). so molality = 0.176 = (moles of solute) / (kg of t-butanol used). 0.176 = (moles of solute) / (0.015 kg of t-butanol) moles = 0.176*0.015 = 0.00264 moles of solute. Our molar mass of the solute would be (.442 g) / (0.00264 moles) = 167.4 g/mol or 167 g/mol. C) To find the freezing point with cyclohexane as the solvent, first find the molality of the solution. molality = (moles of solute) / (kg of cyclohexane) our moles of solute is the same as before, 0.00264 moles. molality = 0.00264 / 0.015 kg cyclohexane = 0.176 m. Use the equation ∆Tf = Kf*m to find the temp change between the freezing pt. of pure cyclohexane and the freezing pt. of the solution. You know what the molality is, and the Kf of cyclohexane is 20 °C/m. ∆Tf = 20*0.176 = 3.52 °C. Now just subtract 3.52 from the freezing point of pure cyclohexane (6.50 °C) and you get the freezing point of the solution: 6.50 - 3.52 = 2.98 °C. Hope this helps!
What is the molarity of 4 grams of NaOH dissolved in 250 ml of solution?
The Molariry = moles of solute/volume of solution in litres ..in order to find the moles from the mass we need to find the molar mass..molar mass for NaOH= ( 22.9+16+1) respectively =39 g/molby using this simple equation we can calculate the moles of Naohmoles= mass/ molar massmoles=4g/39g/mole=o.1 mole.. make sure to cancel the g with g during calculation .now I have the moles and I have the volume I can calculate the molarity, but I have one simple problem that the volume is in (ml )and is supposed to be in (liter).1 litre=1000mlsimple divide what we have over what we want250 ml/1000 liter =0.25 litre yayNow executemolarity= moles/ volumeMolarity =0.1 mole/0.25 liter=0.4 M or mole/liter.. don't forget to write the measurement unit for molarity to get extra grade ..
What is the molarity of solution if 0.1 grams mole of urea is dissolved in 100 grams of water?
100 ml.of the Solution contains= 0.1 G.mole1000 ml.of the Solution contains= 0.1 x1000/100= 1 G.mole.So Molarity of Urea Solution= 1M.(Alternative Method):Molecular Mass if Urea=60100 ml.of Solution contains=0.1 x 60= 6 g1000 ml of Solution contains=60 g.Molarity=Weight in 1Lit/Mole.Wt =60/60=1M
What is the molarity of 2.8 moles of NaCl in 250 mL of solution?
The molarity of a solution is moles per liter. If you have 2.8 moles of NaCl in 250mL, then you have .25 liters (1/4). So, how many moles would you have if you had 1 liter? Multiply your moles in 250mL by 4. There would be 11.2 moles of NaCl in 1 liter. Your molarity is 11.2.
Calculate the molarity of a solution?
Molarity= moles solute/L solute =(0.0345 mol ammonium chloride)/(400.0 ml water) =8.625 =8.6 (because it wants a 2 sig. fig. answer) that's what you said, now I'm going to edit it Molarity= moles solute/L solute =(0.0345 mol ammonium chloride)/(400.0 ml water) --> 0.0345 / .400 L = 0.08625 = 0.086 (because it wants a 2 sig. fig. answer) you divided by mL, not L, so that's why your answer was off
In molarity do we consider the mass of solute in calculation of volume of solution?
Molarity or molar concentration is defined as the number of moles of a solute in a unit volume of solution(usually a litre)......mathematically put, Molarity = (No of moles of solute)/(Unit volume of a solution)......so the answer to your question is no......we don't consider mass of the solute directly.....but the number of moles can be written mathematically as,n = (Mass of the solute)/(Molecular weight of the solute)......However there is a quantity in which mass of the solute is considered directly......It is called Mass concentration and is mathematically put as,Mass concentration = (Mass of the solute)/(Unit Volume of the solution).....
What is the molality and molarity of water?
That's a famous but wrong question. Molarity and molality are terms of concentrations used to find out the concentration of a solute in the solution. You can see that water is neither a solution nor a solvent. However some school or maybe if dumb enough, collage teachers might want you to write this answer-Molarity= moles per litre(Since the specific density of water is 1, ‘molarity and molality of water’ are same)1mole of H2O =18 grams1litre of H2O=1kg of H2Omoles= given weight/molar mass= 1000/18=55.5551 kg/ 1 litre of water(solution(?)) contains 55.55 moles of water(solute(?))Hence molarity and molality of water is 55.55
Calculate the molarity of (M) of 36.5 g HCL in 1.0 L of solution?
Given: mass of solute = 36.5g volume of solution in liters = 1.0L molar mass of HCl = 36.46g/mol Molarity = moles of solute / volume of solution (in liters) moles of solute = mass of solute / molar mass of solute Therefore Molarity = (mass of solute / molar mass of solute) / volume of solution (in liters) Molarity = (36.5g / 36.36g/mol) / 1.0L Molarity = 1.00mol/L = 1.0M Answer: 1.0 M
How do you calculate the molarity of the soln?
This one's pretty tricky... (You're NOT stupid.) Let's start with some relevant definitions: A solution comprises a solute, the thing being dissolved, and a solvent, the thing doing the dissolving. mole fraction = (moles of solute) / (moles solution) [1] molarity = (moles of solute) / (liters of solution) [2] For this conversion, let's use 1 mole of solution to make the calculation easier and the concepts a bit more clear (though it really doesn't matter how much we use). From [1]: 0.0194 = (moles of solute) / 1 moles of solute = 0.0194 moles of solution = (moles of solute) + (moles of solvent) 1 = 0.0194 + (moles of solvent) moles of solvent = 0.9806 The only thing given in the problem to help us with converting to "liters of solution" is the density of the solution (in g/ml). We'll first have to convert the moles of solute and moles of solvent to grams in order to use the density for conversion to liters: g solute = (moles of solute)(molar mass of solute KNO3) molar mass of KNO3 = 39.098 + 14.007 + 3*15.999 g/mol = 101.102 g/mol g solute = (0.0194 mol)(101.102 g/mol) = 1.96 g --- g solvent = (moles of solvent)(molar mass of solvent H2O) molar mass of H2O = 2*1.0079 + 15.999 g/mol = 18.0148 g/mol g solvent = (0.9806 mol)(18.0148 g/mol) = 17.67 g --- g solution = (g of solute) + (g of solvent) g solution = 1.96 + 17.67 g = 19.63 g Now we can convert g solution into L of solution: (19.63 g)(ml / 1.0627 g)(L / 1000 ml) = 0.01847 L From [2]: molarity = (0.0194 mol) / (0.01847 L) = 1.05 M
Calculate the molarity of the NaCl solution when 29.2 g of NaCl is dissolved in 100 mL of water.?
1M NaCl solution is 58.5 g/1000 ml , where molecular wt of NaCl being 58.5 grams , so 29.2 g of NaCl in 1000 ml of water will be equal to : 29.2g /58.5g = 0.499145299 M , so 29.2 g NaCl dissolved in 100 ml would correspond to , 0.499145299 M X 1000 ml/100 ml = 4.99145 M or about 4.99 M