Critical item 11:

1.
Calculate the boiling point for a solution consisting of 10.0 g of ethanol (molar mass = 32 g/mol) and 500 g of water.
     The K_b for water is 0.516 K mol^-1 kg
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2.
Calculate the molecular weight of a compound which when 0.250 g of it is dissolved in 100 g of benzene lowers the melting point by 3.0 K.  The K_f for benzene is 8.0 K mol^-1 kg.
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3
Calculate what weight of ethylene glycol (molar mass = 62 g/mol) is needed to lower the freezing point of 10.0 L of water by 1.0 K.  The K_f for water is 1.86 K mol^-1 kg.
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4
Calculate the boiling point for a solution consisting of 10.0 g of sodium chloride (molar mass = 58.5 g/mol) and 500 g of water.  The K_b for water is 0.516 K mol^-1 kg.
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5
Calculate the molar mass of an ionic compound of the formula AC₃ which when 6.65 g of it is disolved in 100 g of water lowers the melting point by 3.0 K.  The K_f for water is 1.86 K mol^-1 kg.
 
answer

1 ANS

ΔT =  K_bb

n = 10.0 g / 32 g mol^-1

n = 0.313 mol

b = 0.313 mol / 0.500 kg

b = 0.626 mol kg^-1

ΔT = (0.516 K mol^-1 kg) ( 0.626 mol kg^-1 )

ΔT =  0.323 K

T = (100.000 + 0.323) ^oC    (notice for ΔT that ^oC and K are the same)

2 ANS 

ΔT =  K_bb

3.0 K = (8.0 K mol^-1 kg)b

b = 0.375 mol kg^-1

b = n / m_{solvent in kg}

0.375 mol kg^-1 = n / (0.100 kg)

n = 0.0375 mol

M = m / n

M = (250 g)/(0.0375 mol)

M = 6.67 x 10³ g mol^-1
 

3 ANS 

ΔT =  K_bb

1.0 K = (1.86 K mol^-1 kg)b

b = 0.538 mol kg^-1

b = n / m_{solvent in kg}

0.538 mol kg^-1 = n / (10.0 kg)

n = 5.38 mol

M = m / n

62 g mol^-1 = m / (5.38 mol)

m = 333 g

M = m / n

58.5 g mol^-1 = (10.0 g) / n_NaCl

 n_NaCl = 0.171

n_Total  =  2 (0.171 mol )

n_Total  =  0.342 mol

b = n / m_{solvent in kg}

b = 0.342 mol / 0.500 kg

b = 0.684 mol kg^-1

ΔT =  K_bb

ΔT =  K_bb

3.0 K = (1.86 K mol^-1 kg) b

b = 1.61 mol kg^-1      

b = n / m_{solvent in kg}

1.61 mol kg^-1  = n / 0.100 kg

n = 0.161 mol

Note this is of solute or total.  

i = 4

therefore:

n_formal = 0.0403 mol

M = m / n

M  = (6.65 g) / 0.0403 mol

 M  =  165 g mol^-1