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Ad libitum [116K]
3 years ago
7

A solution is made by diluting a standard solution by adding 25 mL of the standard solution to 75 mL of distilled water. What is

the concentration of the resulting solution?
Chemistry
1 answer:
Novay_Z [31]3 years ago
5 0

Answer:

The concentration of the resulting solution is 0.25 or one quarter of the standard solution

Explanation:

Whereby the molar concentration of the standard solution is P molar per liter, we have

25 ml of the standard solution contains

25/1000×P moles of the standard solution such that the resultant solution has a volume of 25 + 75 = 100 ml contains 0.025 P moles of the standard solution

The concentration of a solution is the number of moles contained per liter, which gives

100 ml contains 0.025 P moles

1 liter = 1000 ml will contain 1000/100 × 0.025 = 0.25 P moles of the stanard solution

Therefore, the concentration of the resulting solution is 0.25 or one quarter of the standard solution.

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The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m
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Formula used :  

\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}

where,

\Delta T_f = change in freezing point

T_f^o = temperature of pure ethanol = -117.300^oC

T_f = temperature of solution = -117.431^oC

K_f = freezing point constant of ethanol = 1.99^oC/m

i = van't hoff factor = 1   (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

(-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}

\text{Molar mass of compound}=891.10g/mol

Therefore, the molecular weight of this compound is 891.10 g/mol

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