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kvasek [131]
3 years ago
12

The most concentrated solution from among those listed is: A. 150 mL 0.18M NaOH. B. 75 mL 0.23M KNO3. C. 200 mL 0.15M NaNO3. D.

100 mL 0.25M KCl.
Chemistry
1 answer:
Kryger [21]3 years ago
6 0
The most concentrated solution is the one with the highest molarity. Molarity is calculated by taking the moles of solute/ liters of solution.

The already gave you the molarity, so you do not need to calculate it. The most concentrated solution is the 0.25 M KCl. Keep in mind that the volume they gave you does not effect the molarity of the solution. It just tells us that they have 100 mL of a 0.25 M KCl solution. The 100 mL solution has a molarity of 0.25.

Your correct answer would be D) 100 mL 0.25 M KCl.
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\large \boxed{79 \, \%}

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We have two conditions:

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1. Set up the osmotic pressure condition

Π = cRT, so

\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4)  & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}

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