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jarptica [38.1K]

3 years ago

14

Joe is given a 118 g sample of an impure aqueous solution. The solution does not conduct electricity so he assumes the impurity

is a covalent nonelectrolyte. From density measurements the sample appears to contain 100.0 g of water. Additionally, he was able to freeze the sample at -1.80 ºC. What is the molar mass of the impurity?

1 answer:

DochEvi [55]3 years ago

5 0

Answer: 186 g/mol

Explanation:

Weight of solvent (water)=100 g = 0.1 kg (1 kg=1000 g)

Molar mass of solute (impurity) = ?

Mass of solute (impurity) added = mass of solution - mass of solvent (water) = (118- 100) = 18 g

$\Delta T_f=K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

$\Delta T_f$ = change in freezing point

$K_f$ = freezing point constant for water = $1.86^0C/m$

$\Delta T_f=T_f^0-T_f=(0-(1.80))^0C=1.80^0C$

$1.80=1.86\times frac{18}{M\times 0.1}$

$M=186g/mol$

The molecular mass of the impurity is 186 g/mol.

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Read 2 more answers

A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

$-w=\frac{dm}{dt}$

w is the mass flow

m is the mass of salt

$-v*C=\frac{dm}{dt}$

v is the volume flow

C is the concentration

$C=\frac{m}{V+(6-3)*L/min*t}$

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$-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}$

$-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}$

$-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}$

$-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)$

$-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)$

$m=90kg*[2000L/(2000L+3*L/min*t)]$

a) Initially: t=0

$m=90kg*[2000L/(2000L+3*L/min*0)]=90kg$

b) t=210 min (3.5 hr)

$m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg$

c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

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