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viktelen [127]
3 years ago
9

What is the freezing point of a solution prepared from 45.0 g ethylene glycol (C2H6O2) and 85.0 g H2O? Kf of water is 1.86°C/m.

Chemistry
1 answer:
Andrews [41]3 years ago
6 0

Answer:

T_{sol}=-15.9\°C

Explanation:

Hello,

In this case, we can analyze the colligative property of solutions - freezing point depression - for the formed solution when ethylene glycol mixes with water. Thus, since water freezes at 0 °C, we can compute the freezing point of the solution as shown below:

T_{sol}=T_{water}-i*m*Kf

Whereas the van't Hoff factor for this solute is 1 as it is nonionizing and the molality is:

m=\frac{mol_{solute}}{kg\ of\ water}=\frac{45.0g*\frac{1mol}{62g} }{85.0g*\frac{1kg}{1000g} } =8.54m

Thus, we obtain:

T_{sol}=0\°C+(-8.54m*1.86\frac{\°C}{m} )\\\\T_{sol}=-15.9\°C

Best regards.

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Answer:

m_{MgCO_3}=0.0202molMgCO_3

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

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In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3}  \\\\m_{MgCO_3}=0.0202molMgCO_3

Regards.

7 0
3 years ago
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