The molality of the solution = 17.93 m
<h3>Further explanation</h3>
Given
6.00 L water with 6.00 L of ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)
Required
The molality
Solution
molality = mol of solute/ 1 kg solvent
mol of solute = mol of ethylene glycol
- mass of ethylene glycol :
= volume x density
= 6 L x 1.1132 kg/L
= 6.6792 kg
= 6679.2 g
- mol of ethylene glycol (MW=62.07 g/mol)
=mass : MW
=6679.2 : 62.07
=107.608
6 L water = 6 kg water(ρ= 1 kg/L)
we will use the root mean squared speed of oxygen formula
v= (3RT) / M
T= vM/3R
where
v =1.12x104 m/s; velocity
R= (8.31447 kg m2 s-2 K-1 mol-1) gas constant
M = 0.0319988 kg/mol; molar mass of oxygen
substituting the given values, we will get
T(K)= 14.37K or -258.78 C
Oxygen molecules speed of 1.12x104 m/s is at temperature -258.78C.
The answer is: [D]: " 417 cm³ " .
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Explanation: Use the formula:
V₁ /T₁= V₂ /T₂ ;
in which: V₁ = initial volume = 556 cm³ ;
T₁ = initial temperature = 278 K ;
V₂ = final ("new") temperature = 308 K
T₂ = final ("new:) volume = ?
Solve for "V₂" ;
Since: V₁ /T₁= V₂ /T₂ ;
We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
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V₁ /T₁= V₂ /T₂ ; Multiply EACH side of the equation by "T₂ " :
→ T₂ (V₁ /T₁) = T₂ (V₂ /T₂) ;
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to get:
↔ T₂ (V₂ /T₂) = T₂ (V₁ /T₁) ;
→ V₂ = T₂ (V₁ /T₁) ;
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Now, plug in our known values, to solve for "V₂" ;
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→ V₂ = T₂ (V₁ /T₁) ;
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→ V₂ = 308 K ( 556 cm³ /278 K) ;
→ The units of "K" cancel to "1" ; and we have:
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→ V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note: We will keep the units of volume as: "cm³ "; since all the answer choices given are in units of: "cm³ " ; {that is, "cubic centimeters"}.
→ [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
→ For the "(115,648)" ; round to "3 (three significant figures)" ;
→ "(115,648)" → rounds to: "116,000" ;
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→ (116,000) / (278) = 417.2661870503597122 ;
→ round to 3 significant figures; → "417 cm³ " ;
→ which corresponds with "choice [D]".
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The answer is: [D]: "417 cm³ " .
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