M1V1 = M2V2
M1 = 1.04 * 10-1 = 0.104 M ; V1 = 29.267 mL
M2 =? V2 = 40 mL
M2 = M1V1 / V2 = (0.104 * 29.267) / 40 = 0.0761 M
M = moles / vol (L) ; moles = M * V = 0.0761* 0.04 = 0.00304 moles od tartaric acid present
Weight = molecular weight * moles = 150*0.00304 = 0.4569
If 40 mL wine contains 0.4569 g of tartaric acid, then 1000 mL or 1L will contain 11.42 g. So the concentration expressed in g/L is 11.42.
15) 14.21 g/L of tartaric acid means 14.21 g of tartaric acid in 1000 mL of water. SO per 50 mL itequals to 0.7105 g of tartaric acid.
1000 mL = 14.21 g
50 mL = ?
50*14.21/1000 =0.7105 g o
So 0.7105 g of tartaric acid in 50 mL equal to that concentratoon
A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium hydroxide....
A 20.00 mL wine sample is titrated with a 1.04×10-1M solution of sodium hydroxide. The equivalence point is determined to be 25.926 mL. What is the acidity of the wine expressed as (g tartaric acid / L; MW tartaric acid = 150.085 g/mole)?
The acidity of a wine sample is determined to be 13.08 g tartaric acid/L. What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric flask? (MW tartaric acid = 150.085 g/mole.)
The acidity of a wine sample is 1.05×101, expressed as (g citric acid / L wine; MW citric acid = 192.124 g/mole) . The sample is titrated with a 0.118 M solution of sodium hydroxide. The equivalence point is determined to be 39.740 mL. What is the volume of the wine sample analyzed in mL?
please show all work. thank you
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A 10. equivalence point. L sample of vinegar, an aqueous solution of acetic acid (HC2H02),...