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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?
Chemistry
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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4 0
1 year ago
Calculate the no. of each atoms presents in 10 grams of calcium carbonate.
raketka [301]

Molar Mass of Calcium carbonate:-

\\ \sf\longmapsto CaCO_3

\\ \sf\longmapsto 40u+12u+3(16u)

\\ \sf\longmapsto 52u+48u

\\ \sf\longmapsto 100u

\\ \sf\longmapsto 100g/mol

  • Given Mass=10g

\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}

\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}

\\ \sf\longmapsto No\:of\:moles=0.1mol

Now

\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}

\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}

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Water from a lake changes to a gas state in the process of what?
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3 years ago
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}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-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.

6 0
3 years ago
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