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In the laboratory you dissolve 13.9 g of potassium hydroxide in a volumetric flask and add water to a total volume of 375 . mL. What is the molarity of the solution?
no. Of moles of potassium hydroxide = 13.9/56.1056 = 0.247 molesVolume = 375ml = 0.375LMolarity = no. Of moles/ volume Molarity = 0.247/0.375 = 0.658 M
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When transferring the dissolved solute into the volumetric flask, you should rinse the beaker with solvent to ensure that all the solution is transferred. True or False
True
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X cm³ of 0.15 mol dm³ zinc nitrate solution Zn(NO3)2 is pipetted into a 500 cm³ volumetric flask to produce 500 cm³ of 0.018 mol dm³ zinc nitrate solution. Zn(NO3)2. Determine the value of X.
Molar mass of Zinc nitrate = 189.36 189.36 × 0.018 =3.41×10 = 34.1= 500× 0.15/34.1= 2.2cm3
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In the laboratory you dissolve 17.7 g of ammonium acetate in a volumetric flask and add water to a total volume of 125 mL.
Weight of ammonium acetate = 17.7g No.mole of ammonium acetate = 17.7/77.082 = 0.23 Total volume = 125 ml Molarity of solution = 0.23×1000/125 = 1.8 M
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why should we cool the contents of the conical flask before transferring them into a volumetric flask?
Answer: Because all volumetric flasks are calibrated and graduated for liquids at standard temperature (usually 20 degrees Celcius). This temperature is indicated on the flask.
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<ol><li>Calculate the volume of the original sample needed if you would use a 50 mL volumetric flask, a 100 mL volumetric flask, and a 250 mL volumetric flask to prepare a dilution of 1:25.</li></ol>
Ratio 1:25 means that there have to be 1+25 = 26 equal parts.1) The volume of the original sample needed if we would use a 50 mL volumetric flask is...
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You need to make an aqueous solution of 0.123 M nickel(II) fluoride for an experiment in lab, using a 500 mL volumetric flask. How much solid nickel(II) fluoride should you add?
The amount of substance of NiF2 is: n(NiF2)=V*CM(NiF2) = 0.5 L*0.123 M= 0.0615 mol The mass of NiF2 is: m(NiF2) = M(NiF2)* n(NiF2) = 97 g/mol* 0.0615 mol = 5.9655...
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If you are provided with a 200 ppm solution of quinine, how much would you need to transfer to a 50 mL volumetric flask to prepare a 500 ppb solution? What would its concentration be in mol L-1?
1 ppm = 1000 ppb c_1×V_1=c_2×V_2 200×1000×V_1=500×0.05 V_1=(500×0.05)/(200×1000)=0.000125 (l)=0.125 (ml) 1 ppb = 1×〖10〗^(-6) g/l 500 ppb = 500×〖10〗^(-6) g/l n(C_20 H_24 N_2 O_2 )=(500×〖10〗^(-6))/324,42=1.5×〖10〗^(-6) (mol) c(C_20 H_24 N_2 O_2...
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In the laboratory you dissolve 15.1 g of zinc acetate in a volumetric flask and add water to a total volume of 250 . mL.
What is the molarity of the solution?
_______M
M(Zn(CH3COO)2) = 183.48 g/moln(Zn(CH3COO)2) = m / M = 15.1 / 183.48 = 0.082c(Zn(CH3COO)2) = n / V = 0.082 / 0.25 = 0.328 MAnswer: c(Zn(CH3COO)2) = 0.328 M
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In the laboratory you dissolve 21.3 g of barium acetate in a volumetric flask and add water to a total volume of 375ml what is the molarity of the solution
M (Ba(CH3COO)2) = 255.41 g/moln = m/M n = 21.3/255.41 = 0.083 molC = n/VC = 0.083 mol / 0.375 LC = 0.22 M
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