# chemistry

State how you would prepare each solution:
1) 455g of a solution of C3H5(OH)3 in water in which the mole fraction of glycine is .250
2) 755g of 1.4 molality solution of napathalene C10H8 in benzene C6H6.

My instructor told us to find the amount of grams in order to get the given grams, which confused me, and I really need help and thank you.

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1. What you have written is the formula for glycerine and not glycine.
Let G = grams glycerine
and 455-G = grams H2O
------------------------
n = mols; nG = mols glycerine; nH2O = mols H2O
molar mass glycerine = about 92
molar mass H2O = about 18.
Then [nG/(nG + nH2O)] = xglycerine = 0.25
nG = grams glycerine/92
nH2O = grams H2O = (455-G)/18
Substitute for nG and nH2O in the above equation and solve for grams glycerine which is the only unknown in the equation.

2. A 1.4 m solution of naphthalene in benzene will have 1.4 mols naphthalene in ? kg benzene.

molar mass naphthalene = 128
mols naphthalene = g N/128 which we will call N/128.
m = mols solute/kg solvent
kg benzene = 0.755 - kg N
1.4 = (N/128)
1.4 = (N/128)/kg benzene
1.4 = (N/128)/(0.755 - 0.001N)
Solve for N and I get something like 115 but that isn't exact.
Then benzene = 0.755-0.115 = about 0.64
So 115 g naphthalene in 0.64 kg benzene will give you
(115/128)/0.64 = 1.4m

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